Hc Verma II for Class 11 Science Physics Chapter 27 - Specific Heat Capacities Of Gases

Answer:

No, a gas doesn't have just two specific heat capacities, as the heat capacities depend on the process followed. There are infinite processes; therefore, there can be infinite number of specific heat capacities.

Answer:

Specific heat capacity, $s=\frac{\mathrm{\Delta }Q}{m \mathrm{\Delta }T}$, where $\Delta$Q/m is the heat supplied per unit mass of the substance and $\mathrm{\Delta }$T is the change in temperature produced. At constant temperature, ​$\mathrm{\Delta }$T = 0; therefore, s = infinity. So, we cannot define specific heat capacity at constant temperature.

Answer:

Specific heat capacity, $s=\frac{\mathrm{\Delta }Q}{m \mathrm{\Delta }T}$, where$∆$Q/m is the heat supplied per unit mass of the substance and $∆$T is the change in temperature produced. In an adiabatic process, no heat exchange is allowed; so, $∆$Q = 0 and hence, s = 0. Therefore, in an adiabatic process, specific heat capacity is zero.

Answer:

Yes, a solid also has two kinds of molar heat capacities, C_p and C_v. In a solid, expansion coefficient is quite small; therefore dependence of heat capacity on the process is negligible. So, C_p > C_v with just a small difference, which is not equal to R.

Answer:

In a real gas, as the internal energy depends on temperature and volume, the derived equation for an ideal gas $(\mathrm{d}Q{)}_{\mathrm{p}}=(\mathrm{d}Q{)}_{\mathrm{v}}+nR \mathrm{d}T$ will change to ​$(\mathrm{d}Q{)}_{\mathrm{p}}=(\mathrm{d}Q{)}_{\mathrm{v}}+nR \mathrm{d}T$ + k, where k is the change in internal energy (positive) due to change in volume when pressure is kept constant. So, in the case of a real gas, for n=1 mole (say),
 $$\begin{gathered} C_{\mathrm{p}}-C_{\mathrm{v}}=R+\frac{k}{ \mathrm{d}T} \\ \Rightarrow C_{\mathrm{p}}-C_{\mathrm{v}}>R, \end{gathered}$$
where C_p and C_v are the specific heat capacities at constant pressure and volume, respectively.

Answer:

According to the first law of thermodynamics, change in internal energy, $∆$U is equal to the difference between heat supplied to the gas, $∆$Q and the work done on the gas,​$∆$W, such that $\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$​. In an adiabatic process, $∆$Q = 0 and in an isothermal process, change in temperature, $∆$T = 0. Therefore,
$$\begin{gathered} \mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W \\ \Rightarrow \mathrm{\Delta }Q=n{C}_v\mathrm{\Delta }T+\mathrm{\Delta }W \\ \Rightarrow 0=n{C}_v\left(0\right)+\mathrm{\Delta }W \\ \Rightarrow \mathrm{\Delta }W=0, \end{gathered}$$
where C_v is the heat capacity at constant volume.
This shows that if the process is adiabatic as well as isothermal, no work will be done. So, a process on an ideal gas cannot be both adiabatic and isothermal.

Answer:

In an isothermal process, 
PV = k     ...(i)
On differentiating it w.r.t V, we get
$$\begin{gathered} V\frac{ \mathrm{d}P}{\mathrm{d}V}+P=0 \\ \frac{ \mathrm{d}P}{\mathrm{d}V}=-\frac{P}{V} \\ \frac{ \mathrm{d}P}{\mathrm{d}V}=-\frac{k}{V^2} [\text{U}\mathrm{sing} (\mathrm{i}\left)\right], \text{k = constant} \end{gathered}$$
k = constant
In an adiabatic process, 
$P{V}^{\gamma }=K .... \left(\mathrm{ii}\right)$
On differentiating it w.r.t V, we get
$$\begin{gathered} V^{\gamma }\frac{ \mathrm{d}P}{\mathrm{d}V}+\gamma P{V}^{\gamma -1}=0 \\ \frac{ \mathrm{d}P}{\mathrm{d}V}=-\frac{ \gamma P{V}^{\gamma -1}}{V^{\gamma }} \\ \frac{ \mathrm{d}P}{\mathrm{d}V}=-\frac{\gamma K}{V^{\gamma +1}} [\text{U}\mathrm{sing} (\mathrm{ii}\left)\right], \gamma >1 \text{and} \end{gathered}$$
K is constant
$\gamma \text{and} \frac{\mathrm{d}P}{\mathrm{d}V}$ are the slope of the curve and the ratio of heat capacities at constant pressure and volume, respectively; P  is pressure and V is volume of the system. 
By comparing the two slopes and keeping in mind that $\gamma >1$, we can see that the slope of the P-V diagram is greater for an adiabatic process than an isothermal process.

Answer:

For an isothermal process, PV = K, where P is pressure, V is volume of the system and K is constant. In an isothermal process, a small change in V produces only a small change in p, so as to keep the product constant. On the other hand, in an adiabatic process, $P{V}^{\gamma }=k, \gamma =\frac{C_{\mathrm{P}}}{C_{\mathrm{V}}}$ > 1 is the ratio of  heat capacities at constant pressure and volume, respectively, and k is a constant. In this process, a small increase in volume produces a large decrease in pressure. Therefore, an isothermal process is considered to be a slow process and an adiabatic process a quick process.

Answer:

For two states to be connected by an isothermal process,
$P_1{V}_1=P_2{V}_2$   ...(i)
For the same two states to be connected by an adiabatic process,
$P_1{V_1}^{\gamma } = P_2{V_2}^{\gamma }$   ...(ii)
If both the equations hold simultaneously then, on dividing eqaution (ii) by (i) we get
${V_1}^{\gamma -1} = {V_2}^{\gamma -1}$
Let the gas be monatomic. Then,
$\gamma = \frac{5}{3}$
$$\begin{gathered} \Rightarrow {V_1}^{\frac{2}{3}} = {V_2}^{\frac{2}{3}} \\ \Rightarrow V_1 = V_2 \end{gathered}$$
If this condition is met, then the two states can be connected by an isothermal as well as an adiabatic process.

Answer:

For the molecules of a gas, $\gamma =\frac{C_{\mathrm{p}}}{C_{\mathrm{v}}}=1+\frac{2}{f}$,
where f is the degree of freedom.
$$\begin{gathered} \mathrm{Given}: \gamma =1.29 \\ \Rightarrow 1+\frac{2}{f}=1.29=\frac{9}{7} \\ \Rightarrow \frac{2}{f}=\frac{9}{7}-1 \\ \Rightarrow \frac{2}{f}=\frac{2}{7} \\ \Rightarrow f=7 \end{gathered}$$
Therefore, the molecules of this gas have 7 degrees of freedom.
But in reality, no gas can have more than 6 degrees of freedom.

Answer:

(c) C_A > C_B

According to the first law of thermodynamics, $\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$, where $∆$Q is the heat supplied to the system when $\Delta$W work is done on the system and $∆$U is the change in internal energy produced. Since the temperature rises by the same amount in both processes, change in internal energies are same, i.e. $∆$U_A = $∆$U_B.
But as, $∆$W_A =  $∆$W_B this gives $\Delta$Q_A = 2$\Delta$Q_B.
Now, molar heat capacity of a gas, $C=\frac{\mathrm{\Delta }Q}{n \mathrm{\Delta }T}$, where $∆$Q/n is the heat supplied to a mole of gas and $∆$T is the change in temperature produced. As $∆$Q_A = 2$∆$Q_B, C_A > C_B.

Answer:

c) C_p is slightly greater than C_v

For a solid with a small expansion coefficient, work done in a process will also be small. Thus, the specific heat depends slightly on the process. Therefore, C_p is slightly greater than C_v.

Answer:

(a) p_A < p_B and T_A > T_B

 C_p − C_v  = R for the gas in state A, which means it is acting as an ideal gas in that state, whereas C_p − C_v = 1.08R in state B, i.e. the behaviour of the gas is that of a real gas in that state. To be an ideal gas, a real gas at STP should be at a very high temperature and low pressure. Therefore, P_A < P_B and T_A > T_B where P_A and P_B denotes the pressure and T_A and T_B denotes the temperature of system A and B reepectively.

Answer:

(c) C_p − C_v

For an ideal gas, C_p − C_v = R , where C_v and C_p denote the molar heat capacities of an ideal gas at constant volume and constant pressure, respectively  and R is the gas constant whos value is 8.314 J/K. Therefore, C_p − C_v is a constant. On the other hand, the ratio of these two varies as the atomicity of the gas changes. Also, their sum and product are not constant.

Answer:

(b) 50 calories
It is given that 70 calories of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30° C to 35° C. Also, specific heat at constant pressure,
$$\begin{gathered} C_{\mathrm{p}}=\frac{∆Q}{n ∆T} \\ \Rightarrow C_{\mathrm{p}}=\frac{70}{2\times (35-30)} \\ \Rightarrow C_{\mathrm{p}}=\frac{70}{2\times 5} \\ \Rightarrow C_{\mathrm{p}}=7 \mathrm{calories}-{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1} \end{gathered}$$
For an ideal gas,
$$\begin{gathered} {\mathrm{C}}_{\mathrm{p}}-{\mathrm{C}}_{\mathrm{v}}=\mathrm{R}=8.314 \mathrm{J}-{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\simeq 2 \mathrm{calories} {\mathrm{mol}}^{-1}{\mathrm{K}}^{-1} \\ \Rightarrow C_{\mathrm{v}}=C_{\mathrm{p}}-R \\ \Rightarrow C_{\mathrm{v}}=(7-2) \mathrm{calories}-{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1} \\ \Rightarrow C_{\mathrm{v}}=5 \mathrm{calories}-{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1} \\ \Rightarrow C_{\mathrm{v}}=\frac{∆Q}{n ∆T} \\ \Rightarrow 5=\frac{∆Q}{2\times (35-30)} \\ \Rightarrow ∆\mathrm{Q}=5\times 2\times (35-30) \\ \Rightarrow ∆\mathrm{Q}=5\times 2\times 5 \\ \Rightarrow ∆\mathrm{Q}=50 \mathrm{calories} \end{gathered}$$

Therefore, 50 calories need to be supplied to raise the temperature of 2 moles of gas from 30-35 ^oC at constant volume.

Answer:

(c) C > C_​v

Consider two processes AB and ACB; let W be the work done. C is the molar heat capacity of process AB. Process ACB can be considered as the sum of the two processes, AC and CB. The molar heat capacity of process AC is C_p, as pressure is constant in this process and the molar heat capacity of process CB is C_v, as volume is constant in it. 
Internal energy, U, is a state function, i.e. it doesn't depend on the path followed. Therefore,
 $$\begin{gathered} U_{\mathrm{AB}} = U_{\mathrm{ACB}} \\ {W}_{\mathrm{AB}}>W_{\mathrm{ACB}} \end{gathered}$$
Work done in the p-V diagram is the area enclosed under the curve.
$$\begin{gathered} \Rightarrow W_{\mathrm{AB}}+U_{\mathrm{AB}}>W_{\mathrm{ACB}}+U_{\mathrm{ACB}} \\ \Rightarrow C>C_{\mathrm{v}}+C_{\mathrm{p}} \end{gathered}$$
Molar heat capacity is the heat supplied per mole to change the temperature by a degree Kelvin and according to the first law of thermodynamics, dQ = dU + dW, where dQ is the heat supplied to the system in a process.
$\Rightarrow C>C_{\mathrm{v}}$
_​

Answer:

(d) C = 0.

The defined process is 
$$\begin{gathered} p=\frac{k}{V^{\mathrm{g}}} \\ \Rightarrow p{V}^{\mathrm{g}}=k, \end{gathered}$$
such that the process is adiabatic in which there's no heat supplied to the system, i.e. Q = 0. Molar heat capacity is the amount of heat supplied to the system per mole to produce a degree change in temperature. Also, in an adiabatic process, no heat exchange is allowed. So, molar heat capacity equals zero, i.e. C = 0.

Answer:

(b) 0.5%

Let p and p' be the initial and final pressures of the system and V and V' be the initial and final volumes of the system.  p' is 0.5% more than p and the process is isothernal. So, pV = k = p'V' =  constant. Therefore, 
$$\begin{gathered} pV=p\text{'}V\text{'} \\ \Rightarrow pV=\left(p+\frac{0.5}{100}p\right)V \\ \Rightarrow pV=\frac{100.5}{100}pV\text{'} \\ \Rightarrow V\text{'}=\frac{100}{100.5}V \\ \Rightarrow V\text{'}-V = \frac{100}{100.5}V-V \\ =-\frac{0.5}{100.5} \\ =-0.49\% \end{gathered}$$
So, volume V' decreases by about 0.5% of V.

Answer:

(a) 0.36 %

Let p and p^, be the initial and final pressures of the system and V and V^, be the initial and final volumes of the system. p^, is 0.5% more than p and the process is adiabatic. So,

$$\begin{gathered} p{V}^{\mathit{\gamma }}=p\text{'}V{\text{'}}^{\gamma },\mathit{ }\gamma =1.4=\frac{7}{5} \\ \Rightarrow \frac{V{\text{'}}^{\gamma }}{V^{\gamma }}=\frac{p}{p\text{'}} \\ \Rightarrow {\left(\frac{V\text{'}}{V}\right)}^{\gamma }=\frac{p}{p+{\displaystyle \frac{0.5p}{100}}} \\ \Rightarrow {\left(\frac{V\text{'}}{V}\right)}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}=\frac{100}{100.5} \\ \Rightarrow \frac{V\text{'}}{V}={\left(\frac{100}{100.5}\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}=(\frac{200}{201}{)}^{\frac{5}{7}} \\ \Rightarrow \frac{V\text{'}}{V}=0.99644 \\ \Rightarrow V\text{'}=0.99644V \\ \Rightarrow V\text{'}=V-0.00356V \end{gathered}$$
Therefore, V^, is 0.36 % less than V.

Answer:

(c) p_A < p_B


Let the initial states of samples A and B be i and the final states of samples B and A be f and f', respectively. Let the final volumes of both be V_o. As sample A is expanded through an adiabatic process, its curve in the p-V diagram is steeper than that of sample B, which is expanded through an isothermal process. Therefore, from the p-V diagram, p_A < p_B.

Answer:

(a) T_a < T_b
As sample B is undergoing expansion through an isothermal process, its initial and final temperatures will be same, i.e. T_b. On the other hand, sample A is at the same initial state as B, such that the initial temperature of A is _​T_b and it is expanding through an adiabatic process in which no heat is supplied. Therefore, sample A will expand at the cost of its internal energy and its final temperature will be less than its initial temperature.
This implies that  T_a < T_b.

Answer:

(c) ∆W_a < ∆W_b

In the p-V diagram, the area under the curve w.r.t the V axis is equal to the work done by the system. Since the area under the isotherm is greater than that under the adiabat, the work done by system A is less than that done by system B. Hence,  ∆W_a < ∆W_b.

Answer:

(d) molecular vibrations gradually become effective

Molar specific heat capacity has direct dependence on the degree of freedom of gas molecules. As temperature is increased, the gas molecules start vibrating about their mean position, leading to change (increase) in the degree of freedom and, hence, increasing molar heat capacity.

Answer:

(c) may be very nearly adiabatic
(d) may be very nearly isothermal
Due to sudden compression, the gas did not get sufficient time for heat exchange. So, no heat exchange occurred. Therefore, the process may be adiabatic. For any process to be isothermal, its temperature should remain constant, i.e. pressure and volume should change simultaneously while their product (temperature) should be constant.

Answer:

(d) Q = W

In an isothermal process, temperature of the system stays constant, i.e. there's no change in internal energy. Thus, U = 0, where U denotes the change in internal energy of the system. According to the first law of thermodynamics, heat supplied to the system is equal to the sum of change in internal energy and work done by the system, such that Q = U + W. As U = 0, Q = W.

Answer:

(a) Q = 0
(d) Q ≠ W 
In an adiabatic process, no heat is supplied to the system; so, Q = 0. According to the first law of thermodynamics, heat given to any system is equal to the sum of the change in internal energy and the work done on the system. So, Q = W+U and as Q = 0, W = -U and  Q ≠ W.

Answer:

(c) A is isothermal and B is adiabatic
The slope of an adiabatic process is greater than that of an isothermal process. Since A and B are initiated from the same initial state, both cannot be isothermal or adiabatic, as they would be overlapping. But the curve of process B is steeper than the curve of process A. Hence, A is isothermal and B is adiabatic.

Answer:

(c) The pressures of helium and neon will be the same but that of oxygen will be different.
(d) The temperatures of helium and neon will be the same but that of oxygen will be different.

Adiabatic process is expressed as 
$$\begin{gathered} p{V}^{\gamma }=\mathrm{constant} \mathrm{or} \\ T{V}^{\gamma -1}=\mathrm{constant}, \end{gathered}$$ 
where $\gamma =\frac{C_p}{C_v}$ is the ratio of molar heat capacities at constant pressure and volume. 
We know that $\gamma$ is equal to 1.67 and 1.40 for a monatomic gas and a diatomic gas, respectively. Helium and neon are monatomic gases and oxygen is a diatomic gas. Therefore, changing the state of the gases, i.e. reducing the volume will lead to identical changes in temperature and pressure for helium and neon and that will be different for oxygen. 

Answer:

(a) helium
(b) argon

The temperature of one mole of a gas kept in a container of fixed volume is increased by 1 degree Celsius if 3 calories, i.e. 12.54 J of heat is added to it. So, its molar heat capacity, C_​v = 12.54 J $J{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$, as molar heat capacity at fixed volume is the heat supplied to a mole of gas to increase its temperature by a degree. For a monatomic gas, ​C_{​v $\simeq$$\frac{3}{2}\mathrm{R}=1.5\times 8.314$}= 12.54 JK^-1mol^-1. Among the given gases, only helium and argon are inert and, hence, monoatomic. Therefore, the gas may be helium or argon.

Answer:

(a) argon

The energy of a gas is measured as C_vT. All the four cylinders are at the same temperature but the gases in them have different values of  C_v, such that it is least for the monatomic gas and keeps on increasing as we go from monatomic to tri-atomic. Among the above gases, argon is monatomic, hydrogen and nitrogen are diatomic and carbon dioxide is tri-atomic. Therefore, the energy is minimum in argon.

Answer:

Number of moles of the ideal gas, n = 1 mole
Molecular weight of the gas, W = 20 g/mole
Mass of the gas, m =20 g
Velocity of the vessel, V = 50 m/s
Decrease in K.E. of the vessel = Internal energy gained by the gas
$$\begin{gathered} KE=\frac{1}{2}m(u^2-v^2) \\ KE=\frac{1}{2}\times 20\times {10}^{-3}\times (0-50\times 50) \\ KE=-25 \mathrm{J}=\mathrm{gain} \mathrm{in} \mathrm{internal} \mathrm{energy} \mathrm{of} \mathrm{the} \mathrm{gas} \\ \text{C}\mathrm{hange} \mathrm{in} \mathrm{internal} \mathrm{energy} \mathrm{of} \mathrm{a} \mathrm{gas}=\frac{d}{2}nR(∆T), \\ \mathrm{where} d \mathrm{is} \mathrm{the} \mathrm{degre} \mathrm{of} \mathrm{freedom} \mathrm{of} \mathrm{the} \mathrm{gas} \\ \text{F}\mathrm{or} \text{a} \mathrm{monoatomic} \mathrm{gas}, d=3. \\ \mathrm{So}, 25=\frac{3}{2} nR(∆T) \\ \Rightarrow 25=1\times \frac{3}{2}\times 8.31\times ∆T \\ \Rightarrow ∆T=\frac{50}{(3\times 8.3)}=2 \mathrm{K} \end{gathered}$$

Answer:

Given:
Mass of the gas, m = 5 g
Change in temperature of the system, ∆T = 25 − 15°C = 10°C
Specific heat at constant volume, C_v = 0.172 cal/g -°C
Mechanical equivalent, J = 4.2 J/cal

From the first law of thermodynamics,
dQ =  dU + dW
Now,
$∆$V = 0 (Rigid wall of the container keeps the volume constant)
So, dW = P$∆V$= 0
Therefore,
dQ = dU  (From the first law)
Q = m$c_v$dT = 5 × 0.172 × 10
    = 8.6 cal = 8.6 × 4.2 J
    = 36.12 J
So, change in internal energy of the system is 36.12 J.

Answer:

Given:
Mass of the piston (m) = 50 kg
Adiabatic constant of the gas, γ = 1.4
Area of cross-section of the piston (A) = 100 cm²
Atmospheric pressure (P₀) = 100 kPa
g = 10 m/s² 
Distance moved by the piston , x = 20 cm
Work done by the gas,
$$\begin{gathered} d\mathrm{W}=Pdv \\ \text{The p}\mathrm{ressure} \left(p\right) \mathrm{is} \mathrm{because} \mathrm{of} \mathrm{two} \mathrm{factors}: \text{the}\mathrm{first} \mathrm{is} \mathrm{the} \mathrm{initial} \mathrm{pressure} \mathrm{and} \text{the }\mathrm{second} \mathrm{is} \mathrm{the} \mathrm{weight} \mathrm{of} \mathrm{the} \mathrm{piston}. \\ \mathrm{Therefore}, \\ W=\left(\frac{mg}{\mathrm{A}}+P_0\right)\times Adx \\ W=\left(\frac{50\times 10}{100\times {10}^{-4}}+{10}^5\right)\times 100\times {10}^{-4}\times 20\times {10}^{-4} \end{gathered}$$
W = (5 × 10⁴ + 10⁵) × 20 × 10⁻⁴
W = 1.5 × 10⁵ × 20 × 10⁻⁴
W = 300 J

Hence, nRdT = P$∆V$ = 300
$$\begin{gathered} \Rightarrow d\mathrm{T}=\frac{300}{nR} \\ \mathrm{So}, d\mathrm{Q}=n{\mathrm{C}}_{p}dT=n{c}_p\times \left(\frac{300}{nR}\right) \\ \text{U}\mathrm{sing} C_p-C_v=R and \frac{C_p}{C_v}=\gamma , \\ d\mathrm{Q}=\frac{n\gamma R300}{(\gamma -1)n\mathrm{R}} \\ d\mathrm{Q}=\left(\frac{300\times 1.4}{0.4}\right)=1050 \mathrm{J} \end{gathered}$$

Answer:

Specific heat capacity at constant volume, C_v(H₂) = 2.4 cal/g-°C
Specific heat capacity at constant pressure, C_p(H₂) = 3.4 cal/g-°C
Molecular weight, M = 2 g/mol
Gas constant, R = 8.3 × 10⁷ erg/mol-°C
We know: C_p − C_v = 1 cal/g-°C,
where C_{p and} C_v are molar specific heat capacities.
So, difference of molar specific heat,
 C_p × M − C_v × M = 1 cal/mol-°C

Now, 2 × J = R
⇒ 2 × J = 8.3 × 10⁷ erg/mol-°C
⇒ J = 4.15 × 10⁷ erg/cal

Answer:

Given:
$\left(\frac{{\mathrm{C}}_p}{{\mathrm{C}}_v}\right)=\frac{7}{6}$      
Number of moles of the gas, n = 1 mol
Change in temperature of the gas, ∆T = 50 K

(a) Keeping the pressure constant:
Using the first law of thermodynamics,
dQ = dU + dW
∆T = 50 K and $\gamma =\frac{7}{6}$ 
dQ = dU + dW
Work done, dW = PdV
As pressure is kept constant, work done = P($\mathit{∆}V$)
Using the ideal gas equation PV = nRT,
 P($\mathit{∆}V$) = nR($∆$T)
⇒ dW = nR($∆$T)
At constant pressure,  dQ = nC_p dT
Substituting these values in the first law of thermodynamics, we get
 nC_p dT = dU + RdT
⇒ dU = nC_p dT − RdT
Using $\frac{C_p}{C_V}\mathit{=}\gamma \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{\text{and}}\mathit{ }{C}_P\mathit{-}{C}_V\mathit{=}R, \text{we get}$
$dU=1\times \frac{R\gamma }{(\gamma -1)}\times d\mathrm{T}-Rd\mathrm{T}$
   = 7 RdT − RdT
   = 7 RdT − RdT = 6 RdT
   = 6 × 8.3 × 50 = 2490 J

(b) Keeping volume constant:
Work done = 0
Using the first law of thermodynamics,
dU  = dQ
dU = nC_v dT
$$\begin{gathered} =1\times \frac{R}{\gamma -1}\times d\mathrm{T} \\ =1\times \left(\frac{8.3}{{\displaystyle \frac{7}{6}}-1}\right)\times 50 \end{gathered}$$
= 8.3 × 50 × 6 = 2490 J

(c) Adiabatically, dQ = 0
Using the first law of thermodynamics, we get
dU = − dW
     $$\begin{gathered} =\left[\frac{n\times R}{\gamma -1}(T_1-T_2)\right] \\ =\frac{1\times 8.3}{7/6-1}=(T_2-T_1) \end{gathered}$$
     = 8.3 × 6 × 50 = 2490 J

Answer:

Here,

m = 1.18 g = 1.18×10^-3 kg

ΔQ = 2.0×4.2 J

P = 1.0×10⁶ Pa

V = 1.0×10³ cm³ = 1.0×10^-3 m³

T = 300K

Applying eqn. of state

PV = nRT

=> n = PV/RT

=> n = 1.0×10⁵×1.0×10^-3/(8.314×300)

=> n = 0.04

ΔT = 1⁰C

(ΔH)_v = nC_v ΔT

2.0×4.2 = nC_v×1

=> C_v = 8.4/n= 8.4/0.04

=> C_v = 210

Again we know

C_p – C_v =R

=> C_p = R + C_v

=> C_p = 8.3 + 210

=> C_p = 218.3

Now at constant pressure

(ΔH)_p = nC_p ΔT

=> (ΔH)_p = 218.3×0.04×1 = 8.732 J

In calories

=> (ΔH)_p = 8.732/4.2 = 2.08 cal

Answer:

Initial volume of the gas, V₁ = 100 cm³
Final volume = V₂ = 200 cm³
Pressure = 2 × 10⁵ Pa 
Heat supplied, dQ = 50 J

(a) According to the first law of thermodynamics,
dQ = dU + dW
dW = P$∆V=2\times {10}^5\times (200-100)\times {10}^{-6}=20$
⇒ 50 = dU + 2 × 10
⇒ dU = 30 J

(b) For a monatomic gas,
U = $\frac{3}{2}n$RT
 30 = n × $\frac{3}{2}$ × 8.3 × 300

$\Rightarrow n=\frac{2}{(83\times 3)}=\frac{2}{249}=0.008$

(c)  Also,
dU = nC_vdT
$\Rightarrow C_{\mathit{v}}\mathit{=}\frac{\mathit{d}\mathit{U}}{\mathit{n}\mathit{d}\mathit{T}}=\frac{30}{0.008\times 300}=12.5$
C_p = C_v + R = 12.5 + 8.3 = 20.8 J/mol-K

(d) C_v = 12.5 J/mol-K

Answer:

Given:
Amount of heat given to the gas = Q
So, ∆Q = Q
Work done by the gas, ∆W = $\frac{\mathrm{Q}}{2}$
From the first law of thermodynamics,
∆Q = ∆W + ∆U
$\Rightarrow ∆U=\mathrm{Q}-\frac{\mathrm{Q}}{2}=\frac{\mathrm{Q}}{2}$
For a monoatomic gas,
∆U = $\frac{3}{2}$ nRdT
$\Rightarrow \frac{\mathrm{Q}}{2}=ndT\times \frac{3}{2}R$
$\Rightarrow$Q = 3nRdT

Again, for expansion at constant pressure,
Q = nC_pdT,
where C_p is the molar heat capacity at constant pressure.
So, 3RndT = nC_pdT
⇒ C_p = 3R

Answer:

Relation between pressure and volume of a gas is P = kV.

Ideal gas equation is PV = nRT.
$$\begin{gathered} \Rightarrow \frac{nRT}{V}=kV \\ \Rightarrow nRT=k{V}^2 \end{gathered}$$  

For simplicity, take the number of moles of  a gas, n = 1.
⇒ RdT = 2 kVdV
$\Rightarrow \frac{RdT}{2kV}=dV$

From the first law of thermodynamics,
dQ = dU + dW
⇒ n$C_p$dT = C_v dT + PdV
$$\begin{gathered} \Rightarrow n{C}_{p}dT =\mathit{ }{C}_v\mathit{ }dT+\frac{PRdT}{\mathit{2}kV} \\ \Rightarrow 1\times C_p=\mathit{ }{C}_v+\frac{PR}{2kV} \\ \therefore C_p=C_v+\frac{R}{2} \end{gathered}$$     

Answer:

As the process has specific heat capacity zero, the process is essentially an adiabatic process.

Answer:

For the first ideal gas,
C_p1 = specific heat at constant pressure
C_v1 = specific heat at constant volume
n₁ = number of moles of the gas
$\frac{C_{{\mathrm{p}}_1}}{C_{{\mathrm{V}}_1}}$ = γ and C_p1 − C_v1 = R
⇒ γC_v1 − C_v1 = R
⇒ C_v1 (γ − 1) = R
$$\begin{gathered} \Rightarrow {\mathrm{Cv}}_1=\frac{R}{(\mathrm{\gamma }-1)} \\ C{\mathrm{p}}_1=\mathrm{\gamma }\frac{R}{(\mathrm{\gamma }-1)} \end{gathered}$$

For the second ideal gas,
C_p2 = specific heat at constant pressure
C_v2 = specific heat at constant volume
  n₂ = number of moles of the gas
$\frac{{\mathrm{C}}_{{\mathrm{p}}_2}}{{\mathrm{C}}_{{\mathrm{v}}_2}}$ = γ and C_p2 − C_v2 = R
⇒ γC_v2 − C_v2 = R
⇒ C_v2 (γ − 1) = R
$$\begin{gathered} \Rightarrow {\mathrm{Cv}}_2=\frac{R}{(\mathrm{\gamma }-1)} \mathrm{and} \\ {\mathrm{Cp}}_2=\mathrm{\gamma }\frac{R}{(\mathrm{\gamma }-1)} \end{gathered}$$
Given:
n₁ = n₂ = 1 : 2
dU₁ = nC_v1dt
dU₂= 2nC_v2dT

When the gases are mixed,
nC_v1dT + 2nC_v2dT = 3nC_vdT
$$\begin{gathered} {\mathrm{C}}_{\mathrm{v}}=\frac{{\mathrm{C}}_{v_1}+2{\mathrm{C}}_{v_2}}{3} \\ =\frac{{\displaystyle \frac{R}{(\mathrm{\gamma }-1)}}+{\displaystyle \frac{2R}{(\mathrm{\gamma }-1)}}}{3} \\ =\frac{3R}{(\mathrm{\gamma }-1)3}=\frac{R}{\mathrm{\gamma }-1} \end{gathered}$$

Hence, Cp / Cv in the mixture is γ.

Answer:

Specific heat at constant pressure of helium, C_p' = 2.5 R
Specific heat at constant pressure of hydrogen, C_p" = 3.5 R
Specific heat at constant volume of helium, C_v' = 1.5 R
Specific heat at constant volume of hydrogen, C_v" = 2.5 R
n₁ = n₂ = 1 mol
_{$dU\mathit{=}n{C}_{v}dT$}

For the mixture of two gases,
$d{U}_1+d{U}_2=1 \mathrm{mol}$
[n₁ + n₂ ]  C_vdT = n₁C'_vdT + n₂C"_vdT,
where $C_V$ is the heat capacity of the mixture.
$$\begin{gathered} \Rightarrow C_v\mathit{=}\frac{n_{1}C{\mathit{\text{'}}}_v\mathit{+}{n}_{2}C{\mathit{"}}_v}{n_1\mathit{+}{n}_{\mathit{2}}} \\ =\frac{1.5R+2.5R}{2}=2R \end{gathered}$$
Cp = Cv + R = 2R + R = 3R
$\mathrm{\gamma }=\frac{\mathrm{C}p}{\mathrm{C}v}=\frac{3R}{2R}=1.5$

Answer:

Given:
Number of moles of the gas, $n=0.5 \mathrm{mol}$



$$\begin{gathered} R=\frac{25}{3} \mathrm{J}/\mathrm{mol}-\mathrm{K} \\ \gamma =\frac{5}{3} \end{gathered}$$

(a) Temperature at a = $T_a$
$P_{\mathit{a}}{V}_{\mathit{a}}\mathit{=}nR{T}_{\mathit{a}}$
$\Rightarrow Ta\mathit{=}\frac{PaVa}{nR}=\frac{100\times {10}^3\times 5000\times {10}^{-6}}{0.5\times {\displaystyle \frac{25}{3}}}=120 \mathrm{K}$

Similarly, temperature at b,
$$\begin{gathered} T_b\mathit{=}\frac{P_b{V}_b}{nR} \\ {T}_b=\frac{100\times {10}^3\times 10000\times {10}^{-6}}{0.5\times {\displaystyle \frac{25}{3}}} \\ {T}_b=240 K \end{gathered}$$
 Similarly, temperature at c  is 480 K and at d  is 240 K.

(b) For process ab,
dQ = nc_pdT
[Since ab is isobaric]
$$\begin{gathered} dQ=\frac{1}{2}\times \frac{R\gamma }{\gamma \mathit{-}\mathit{1}}\mathit{(}{T}_b\mathit{-}{T}_a) \\ dQ=\frac{1}{2}\times \frac{{\displaystyle \frac{25\times 5}{3\times 3}}}{{\displaystyle \frac{5}{3}}-1}\times (240-120) \\ dQ=\frac{1}{2}\times \frac{125}{9}\times \frac{3}{2}\times \left(120\right) \\ dQ=1250 \mathrm{J} \end{gathered}$$

For line bc, volume is constant. So, it is an isochoric process.
dQ = dU + dW
[dW = 0, isochoric process]
dQ = dU = nC_vdT
dQ= nC_v ($T_c-T_b)$
$$\begin{gathered} dQ=\frac{1}{2}\times \frac{\left({\displaystyle \frac{25}{3}}\right)}{\left[\left({\displaystyle \frac{5}{3}}\right)-1\right]}\times \left(240\right) \\ dQ=\frac{1}{2}\times \frac{25}{3}\times \frac{3}{2}\times 240=1500 \mathrm{J} \end{gathered}$$

(c) Heat liberated in cd (isobaric process),
dQ = − nC_pdT

$$\begin{gathered} dQ=-\frac{1}{2}\times \frac{\gamma R}{\mathit{(}\gamma \mathit{-}\mathit{1}\mathit{)}}\mathit{\times }\mathit{(}{T}_d\mathit{-}{T}_c) \\ dQ=-\frac{1}{2}\times \frac{125}{9}\times \frac{3}{2}\times (240-480) \\ dQ=-\frac{1}{2}\times \frac{125}{6}\times 240=2500 \mathrm{J} \end{gathered}$$

Heat liberated in da (isochoric process),
$\Rightarrow$dQ = dU
Q= −nC_vdT
$$\begin{gathered} dQ=-\frac{1}{2}\times \frac{R}{\gamma \mathit{-}\mathit{1}}\left(Ta\mathit{-}Td\right) \\ dQ=-\frac{1}{2}\times \frac{25}{2}\times (120-240) \\ dQ=\frac{25}{4}\times 120=750 \mathrm{J} \end{gathered}$$

Answer:

(a) For line ab, volume is constant.
So, from the ideal gas equation,
$$\begin{gathered} \frac{P_{\mathit{1}}}{T_{\mathit{1}}}\mathit{=}\frac{P_{\mathit{2}}}{T_{\mathit{2}}} \\ \Rightarrow \frac{100}{300}=\frac{200}{{\mathrm{T}}_2} \end{gathered}$$



$T_{\mathit{2}}=\left(\frac{200\times 300}{100}\right)=600 \mathrm{K}$

For line bc, pressure is constant.
$$\begin{gathered} \mathrm{So}, \frac{{\mathit{V}}_{\mathit{1}}}{{\mathit{T}}_{\mathit{1}}}\mathit{=}\frac{{\mathit{V}}_{\mathit{2}}}{{\mathit{T}}_{\mathit{2}}} \\ \Rightarrow \frac{100}{600}=\frac{150}{{\mathrm{T}}_2} \\ \Rightarrow {\mathrm{T}}_2 \left(\frac{600\times 150}{100}\right)=900 \mathrm{K} \end{gathered}$$

(b)
As process ab is isochoric, $W_{ab}=0$.
During process bc,
P = 200 kPa

The volume is changing from 100 to 150 cm³ .
Therefore, work done = 50 × 10⁻⁶ × 200 × 10³ J
                                    = 10 J

(c) For ab (isochoric process), work done = 0.
From the first law,
dQ = dU = nC_vdT
$\Rightarrow$Heat supplied = nC_vdT
Now,
$$\begin{gathered} Q_{ab}\mathit{=}\left(\frac{PV}{RT}\right)\mathit{\times }\left(\frac{R}{\gamma \mathit{-}\mathit{1}}\right)\mathit{\times }dT \\ =\frac{200\times {10}^3\times 100\times {10}^{-6}\times 300}{600\times 0.67} \\ =14.925 (\therefore \mathrm{\gamma }=1.67) \\ \text{F}\mathrm{or} \mathrm{bc} (\mathrm{isobaric} \mathrm{process}): \\ \mathrm{Heat} \mathrm{supplied} \mathrm{in} \mathrm{bc}=n{C}_{p}dT\mathit{ }\mathit{ }\mathit{ }\mathit{ }\left(C_p\mathit{=}\frac{\gamma R}{\gamma \mathit{-}\mathit{1}}\right) \\ \mathit{=}\frac{PV}{RT}\mathit{\times }\frac{\gamma R}{\gamma \mathit{-}\mathit{1}}\mathit{\times }dT \\ =\frac{200\times {10}^3\times 150\times {10}^{-6}}{600\times 0.67}\times 300 \\ =10\times \frac{1.67}{0.67}=\frac{16.7}{0.67}=24.925 \end{gathered}$$

(d) dQ = dU + dW
Now,
dU = dQ − dW
     = Heat supplied − Work done
     = (24.925 + 14.925) − 10
     = 39.850 − 10 = 29.850 J.

Answer:

For Joly's differential steam calorimeter,
$C_v\mathit{=}\frac{m_{\mathit{2}}L}{m_{\mathit{1}}\mathit{ }\mathit{(}{\theta }_2\mathit{-}{\theta }_1\mathit{)}}$,
where
m₂ = mass of steam condensed
m₂ = 0.095 g
Latent heat of vapourization, L = 540 cal/g = 540 × 4.2 J/g
m₁ = mass of gas present
m₁ = 3 g
Initial temperature, θ₁ = 20°C
Final temperature, θ₂ = 100°C
$\Rightarrow C_v=\frac{0.095\times 540\times 4.2}{3\times (100-20)}$
       = 0.89 = 0.9 J/g-K

Answer:

Given,
γ = 1.5
Since the process is adiabatic, PV^γ = constant.

(a) P₁V₁^γ = P₂V₂^γ
Given, V₁ = 4 L
           V₂ = 3 L
We need to find $\frac{P_2}{P_1}$.
$$\begin{gathered} \Rightarrow \frac{{\mathit{P}}_{\mathit{2}}}{{\mathit{P}}_{\mathit{1}}}\mathit{=}{\left(\frac{{\mathit{V}}_{\mathit{1}}}{{\mathit{V}}_{\mathit{2}}}\right)}^{\mathit{\gamma }} \\ ={\left(\frac{4}{3}\right)}^{1.5}=1.5396=1.54 \end{gathered}$$

(b) Also, for an adiabatic process,
TV^γ−1 = constant
T₁V₁^γ−1 = T₂V₂^γ−1
$\Rightarrow \frac{T_{\mathit{2}}}{T_{\mathit{1}}}\mathit{=}{\left(\frac{V_{\mathit{1}}}{V_{\mathit{2}}}\right)}^{\gamma \mathit{-}\mathit{1}}={\left(\frac{4}{3}\right)}^{0.5}=1.154$

Answer:

Initial pressure of the gas, P₁ = 2.5 × 10⁵ Pa
Initial temperature, T₁ = 300 K
Initial volume, V₁ = 100 cc

(a) For an adiabatic process,
P₁V₁^γ = P₂V₂^γ
⇒ 2.5 × 10⁵ × V^1.5 = ${\left(\frac{\mathrm{V}}{2}\right)}^{1.5}$ × P₂
⇒ P₂ = 7.07 × 10⁵
         = 7.1 × 10⁵ Pa

(b) Also, for an adiabatic process,
T₁V₁^γ−1 = T₂V₂^γ−1
⇒ 300 × (100)^1.5−1 = T₂ × ${\left(\frac{100}{2}\right)}^{1.5-1}$
= T₂ × (50)^1.5−1
⇒ 300 × 10 = T₂   × 7.07
⇒ T ₂  = 424.32 K = 424 K

(c) Work done by the gas in the process,
$$\begin{gathered} W\mathit{=}\frac{nR}{\mathit{(}\gamma \mathit{-}\mathit{1}\mathit{)}}\mathit{ } [T_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}] \\ \mathit{=}\frac{P_{\mathit{1}}{V}_{\mathit{1}}}{T\mathit{(}\gamma \mathit{-}\mathit{1}\mathit{)}}\mathit{ }\mathit{ }[T_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}] \\ =\frac{2.5\times 10}{300\times 0.5}\times (-124) \\ = -20.67 \approx -21 \mathrm{J} \end{gathered}$$

Answer:

Given:
For air, γ = 1.4
Initial temperature of air, T₁ = 20°C = 293 K
Initial pressure, P₁ = 2 atm
Final pressure, P₂ = 1 atm
The bursting of the tyre is an adiabatic process. For an adiabatic process,

P₁^1−γ × T₁^1−γ = P^1−γ × T₂^γ
 (2)^1−1.4 × (293)^1.4 = (1)^1−1.4 × T₂^1.4
⇒ (2)^−0.4 × (293)^1.4 = T₂^1.4
⇒ 2153.78 = T₂^1.4
⇒ T₂ = (2153.78)^1/1.4
         = 240.3 K

Answer:

Initial pressure of the gas, P₁ = 100 kPa
Initial volume of the gas,V₁ = 400 cm³
                                             = 400 × 10⁻⁶ m³
Initial temperature of the gas, T₁ = 300 K
$\gamma =\frac{\mathrm{C}p}{\mathrm{C}v}=1.5$

(a) The gas is suddenly compressed to volume, V₂ = 100 cm³ .
So, this is an adiabatic process.
For an adiabatic process,
P₁V₁^γ = P₂V₂^γ
⇒ 10⁵ × (400)^1.5 = P₂ (100)^1.5 
⇒ P₂ = 10⁵(4)^1.5 = 800 kPa
  
Also,
T₁V^γ−1 = T₂V₂^γ−1
⇒ 300 × (400)^1.5−1 = T₂ (100)^1.5−1
⇒ 300 × (400)^0.5 = T₂ (100)^0.5  
⇒ T₂ = 600 K

(b) If the container is slowly compressed, the heat transfer is zero, even thought the walls are adiabatic.
Thus, the values remain same. Thus,
P₂ = 800 kPa
T₂ = 600 K

Answer:

Given:
For the gas, $\frac{\mathit{C}\mathit{p}}{\mathit{C}\mathit{v}}\mathit{=}\gamma$
Initial pressure of the gas = P₀ 
Initial volume of the gas = V₀

(a)
(i) As the gas is slowly compressed, its temperature will remain constant.
For isothermal compression,
P₁V₁ = P₂V₂
So, P₀V₀ = P₂ $\frac{{\mathrm{V}}_0}{2}$ ⇒ P₂ = 2P₀

(ii) Sudden compression means that the gas could not get sufficient time to exchange heat with its surroundings. So, it is an adiabatiac compression.
So, for adiabatic compression,
P₁V₁^γ = P₂V₂^γ
 $$\begin{gathered} \text{Or} 2P_0{\left(\frac{V_0}{2}\right)}^{\gamma }={\mathrm{P}}_2{\left(\frac{V_0}{4}\right)}^{\mathrm{\gamma }} \\ \Rightarrow P_2=\frac{{V_{\mathit{0}}}^{\gamma }}{2^{\mathrm{\gamma }}}\times 2P_0\times \frac{4^{\mathrm{\gamma }}}{{V_{\mathit{0}}}^{\gamma }} \end{gathered}$$
= 2^γ × 2P₀ = P₀2^γ+1

(b)
(i) Adiabatic compression:
P₁V₁^γ = P₂V₂^γ
 P₀V₀^γ = P'${\left(\frac{V_0}{2}\right)}^{\gamma }$
⇒ P' = P₀2^γ

(ii) Isothermal compression:
P₁V₁ = P₂V₂
$2^{\gamma }{P}_0\times \frac{V_0}{2}=P"\left(\frac{V_0}{2}\right)$
P" = P₀2^γ × 2
⇒ P" = P₀2^γ+1  

Answer:

(a) Given,
Initial pressure of the gas = p₀
Initial volume of the gas = $V_0$
For an isothermal process,
PV = constant
So, P₁V₁ = P₂V₂
$P_{\mathit{2}}\mathit{=}\frac{{\mathit{P}}_{\mathit{0}}{\mathit{V}}_{\mathit{0}}}{\mathit{\left(}{\displaystyle \frac{P_0}{2}}\mathit{\right)}}=2{\mathrm{V}}_0$
For an adiabatic process, P₃ = $\frac{P_0}{4}$, V₃ = ?
P₂V₂^γ = P₃V₃^γ
$$\begin{gathered} \Rightarrow {\left(\frac{V_3}{V_2}\right)}^{\gamma }\mathit{=}\left(\frac{P_2}{P_3}\right) \\ \mathit{\Rightarrow }{\left(\frac{V_3}{V_2}\right)}^{\gamma }\mathit{=}\left(\frac{{\displaystyle \frac{P_0}{2}}}{{\displaystyle \frac{P_0}{4}}}\right)=2 \\ \mathit{\Rightarrow }\frac{V_3}{V_2}=2^{\frac{1}{\gamma }} \\ \therefore V_3\mathit{=}{V}_{2\mathit{ }}{\mathit{2}}^{\frac{\mathit{1}}{\gamma }}=2V_0 2^{\frac{\mathit{1}}{\gamma }} \\ \mathit{=}{2}^{\frac{\gamma \mathit{+}1}{\gamma }}{V}_0 \end{gathered}$$

(b) P₁V₁^γ = P₂V₂^γ
$$\begin{gathered} \text{Or} \left(\frac{V_2}{V_1}\right)={\left(\frac{P_1}{P_2}\right)}^{\frac{1}{\gamma }} \\ \Rightarrow V_2=V_0 2^{\frac{1}{\gamma }} \end{gathered}$$

Again, for an isothermal process,
P₂V₂ = P₃V₃
$$\begin{gathered} \therefore \mathit{ }{V}_{\mathit{3}}\mathit{=}\frac{P_{\mathit{2}}{V}_{\mathit{2}}}{P_{\mathit{3}}}={22}^{\frac{1}{\gamma }}{V}_0 \\ =2^{\frac{\mathrm{\gamma }+1}{\mathrm{\gamma }}}{V}_0 \end{gathered}$$

Answer:

The ideal gas equation is
PV = nRT
Given, $P_1$ = 150 kPa = 150 × 10³ Pa
$V_1$ = 150 cm³ = 150 × 10⁻⁶ m³
$T_1$= 300 K

(a)
$$\begin{gathered} n\mathit{=}\frac{PV}{RT}=9.036\times {10}^{-3} \\ n=0.009 \end{gathered}$$

(b)
$$\begin{gathered} \frac{C_p}{C_v}\mathit{=}\gamma \mathit{,}\mathit{ }{C}_p\mathit{-}{C}_v\mathit{=}R \\ \text{So}\mathit{,}\mathit{ }{C}_v\mathit{=}\frac{R}{\gamma \mathit{-}\mathit{1}}=\mathit{2}R=\frac{8.3}{0.5}=16.6 \mathrm{J}/\mathrm{mole}-\mathrm{K} \end{gathered}$$

(c) Given,
P₁ = 150 kPa = 150 × 10³ Pa
P₂ = ? V₁ = 150 cm³
= 150 × 10⁻⁶ m³
γ = 1.5
V₂ = 50 cm³ = 50 × 10⁻⁶ m³,
T₁ = 300 K
T₂ = ?
Since the process is adiabatic, using the equation of an adiabatic process,we get
P₁V₁^γ = P₂V₂^γ
⇒ 150 × 10³ × (150 × 10⁻⁶)^γ = P₂ × (50 × 10⁻⁶)^γ
$\Rightarrow P_2=150\times {10}^3\times \frac{(150\times {10}^{-6}{)}^{1.5}}{(50\times {10}^{-6}{)}^{1.5}}$
P₂ = 150000 × (3)^1.5
P₂ = 779.422 × 10³ Pa
P₂ = 780 kPa
Again,
P₁^1−γ T₁^γ = P₁^1−γ T₂^γ
⇒ (150 × 10³)^1−1.5 × (330)^1.5 = (780 × 10³)^1−1.5 × T₂^1.5
⇒ T₂^1.5 = (150 × 10³)^1−1.5 × (300)^1.5 × 300^1.5
T₂^1.5 = 11849.050
⇒ T₂ = (11849.050)^1/1.5
   T₂ = 519.74 = 520 K

(d) dQ = dW + dU
Or dW = −dU  [ Since dQ = 0 in an adiabatic process]
dW = −nC_vdT
dW = −0.009 × 16.6 × (520 − 300)
dW = −0.009 × 16.6 × 220
dW = −32.87 J $\approx$ −33 J

(e)
dU = nC_vdT
dU = 0.009 × 16.6 × 220 $\approx$ 33 J

Answer:

There are three gases A, B and C.
It is given that initially,
V_A = V_B = V_C    and     T_A = T_B = T_C .

For A, the process is isothermal and for an isothermal process, PV = constant.
P_AV_A = P'_A2V_A
$\Rightarrow P{\mathit{\text{'}}}_{\mathrm{A}}\mathit{=}{P}_{\mathrm{A}}\mathit{\times }\frac{1}{2}$

For B, the process is adiabatic. So,
P_BV_B^γ = P_B'(2V_B)^γ
$\Rightarrow P_{\mathrm{B}}\text{'}\mathit{=}\frac{P_{\mathrm{B}}}{2^{1.5}}$

For C, the process is isobaric, which implies that the pressure will remain constant.
So, using the ideal gas equation
P$=\frac{\mathit{n}\mathit{R}\mathit{T}}{\mathit{V}}$, we get
$\frac{V_{\mathrm{C}}}{T_{\mathrm{C}}}\mathit{=}\frac{V{\mathit{\text{'}}}_{\mathrm{C}}}{T{\mathit{\text{'}}}_{\mathrm{C}}}\mathit{\Rightarrow }\frac{V_{\mathrm{C}}}{T_{\mathrm{C}}}\mathit{=}\frac{2V_{\mathrm{C}}}{T{\mathit{\text{'}}}_{\mathit{C}}}$
⇒ T_C' = 2T_C

As the final pressures are equal,
$$\begin{gathered} \frac{P_{\mathit{A}}}{2}=\frac{P_{\mathit{B}}}{2^{1.5}}=P_{\mathit{C}} \\ \Rightarrow P_{\mathit{A}}\mathit{ }\mathit{:}\mathit{ }{P}_{\mathit{B}}\mathit{ }\mathit{:}\mathit{ }{P}_{\mathit{C}}=2 : 2^{1.5} : 1 \end{gathered}$$

Answer:

Let,
Initial pressure of the gas = P₁
Initial volume of the gas = V₁ 
Final pressure of the gas= P₂ 
Final volume of the gas = V₂ 
Given, V₂ = 2 V₁, for each case.
In an isothermal expansion process,
work done$=nR{T}_1\mathit{ }\ln \frac{V_2}{V_1}$
Adiabatic work done,
$W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}$
It is given that same work is done in both cases.
So,
$nR{T}_1\mathit{ }\ln \mathit{\left(}\frac{V_2}{V_1}\mathit{\right)}\mathit{=}\frac{P_1{V}_1\mathit{-}{P}_2{V}_2}{\gamma \mathit{-}1} ...\left(1\right)$
In an adiabatic process,
$P_2=P_1{\left(\frac{V_1}{V_2}\right)}^{\gamma }=P_1{\left(\frac{1}{2}\right)}^{\gamma }$
From eq (1),
$nR{T}_1\mathit{ }\ln \mathit{ }2\mathit{=}\frac{P_1{V}_1\left(1-{\displaystyle \frac{1}{2^{\mathrm{\gamma }}}}\times 2\right)}{\gamma \mathit{-}1}$
and nRT₁ = P₁V₁
$\mathrm{So}, \ln 2=\frac{1-{\displaystyle \frac{1}{2^{\mathrm{\gamma }}}}.2}{\mathrm{\gamma }-1}$
Or (γ − 1) ln 2 = 1 − 2^1−γ

Answer:

Given:
γ = 1.5
T = 300 K
Initial volume of the gas, V₁ = 1 L
Final volume, V₂ = $\frac{1}{2}$ L

(a) The process is adiabatic because volume is suddenly changed; so, no heat exchange is allowed.
P₁V₁^γ = P₂V₂^γ
$$\begin{gathered} \text{Or} P_2=P_1{\left(\frac{V_1}{V_2}\right)}^{\gamma }=P_1(2{)}^{\gamma } \\ \frac{P_2}{P_1}=2^{1.5}=2\sqrt{2} \end{gathered}$$

(b) P₁ = 100 kPa = 10⁵ Pa
and P₂ = $2\sqrt{2}$ × 10⁵ Pa
Work done by an adiabatic process,
$$\begin{gathered} W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1} \\ W=\frac{{10}^5\times {10}^{-3}-2\sqrt{2}\times {10}^5\times {\displaystyle \frac{1}{2}}\times {10}^{-3}}{1.5-1} \\ W=-82 \mathrm{J} \end{gathered}$$

(c) Internal energy,
dQ = 0, as it is an adiabatic process.
⇒ dU = − dW = − (− 82 J) = 82 J

(d)
Also, for an adiabatic process,
T₁V₁^γ−1 = T₂V₂^γ−1
T₂ = T₁${\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$
= 300 $\times$ (2)^0.5
= 300 × $\sqrt{2}$ × = 300 × 1.4142
T₂ = 424 K

(e) The pressure is kept constant.
The process is isobaric; so, work done = P$∆V$=nRdT.
Here, $n=\frac{PV}{RT}=\frac{{10}^5\times {10}^{-3}}{R\times 300}=\frac{1}{3R}$
So, work done = $\frac{1}{3R}\times R\mathit{\times }(300-424)=-41.4 \mathrm{J}$
As pressure is constant,

 $$\begin{gathered} \frac{V_1}{T_1}=\frac{V_2}{T_2} ...\left(1\right) \\ {V}_1=V_2\frac{T_1}{T_2} \end{gathered}$$

(f)Work done in an isothermal process,
$$\begin{gathered} W=nRT \ln \frac{V_2}{V_1} \\ =\frac{1}{3R}\times R\times T\times \mathrm{In} \left(2\right) \end{gathered}$$
= 100 × ln 2 = 100 × 1.039
= 103 J

(g) Net work done (using first law of thermodynamics)
= − 82 − 41.4 + 103
= − 20.4 J

Answer:

Given:
γ = 1.5
For an adiabatic process, TV^γ−1 = constant .
So, T₁ V₁^γ−1 = T₂ V₂^γ−1        
As it is an adiabatic process and all the other conditions are same, the above equation can be applied.
In the new position, the slid is dividing the tube in the ratio 3:1.
So, if the total volume is V, then one side will occupy a volume of$\frac{3}{4}V$ and the other side will occupy $\frac{V}{4}$.
So, $T_1\times {\left(\frac{3v}{4}\right)}^{\gamma -1}=T_2\times {\left(\frac{v}{4}\right)}^{\gamma -1}$

(i)   (ii)

$$\begin{gathered} \Rightarrow T_1\times {\left(\frac{3v}{4}\right)}^{1.5-1}=T_2\times {\left(\frac{v}{4}\right)}^{1.5-1} \\ \Rightarrow \frac{T_1}{T_2}=\frac{\sqrt{3}}{1} \end{gathered}$$

Answer:

Given:
Volume of gas in each vessel, V = 200 cm³
Specific heat at constant volume of the gas, C_v = 12.5 J/mol-K
Initial temperature of the gas, T = 300 K  
Initial pressure of the gas, P = 75 cm of Hg

(a) Using the ideal gas equation, number of moles of gases in each vessel,

$$\begin{gathered} 75 \mathrm{cm} \mathrm{of} \mathrm{Hg} = 99991.5 \mathrm{N}/{\mathrm{m}}^2 \\ n\mathit{=}\frac{PV}{RT} \\ =\frac{99991.5\times 200\times {10}^{-6}}{8.3\times 300} \\ =8031.4\times {10}^{-6} \\ =0.008 \end{gathered}$$

(b) Heat is supplied to the gas, but dV is zero as the container has rigid walls.
      So, dW = $P\mathit{∆}V$ = 0

From first law of thermodynamics,
dQ = dU
⇒ 5 = nC_vdT
⇒ 5 = 0.008 × 12.5 × dT
⇒ dT = 50 for A
$\frac{\mathit{P}}{\mathit{T}}=\frac{P_{\mathrm{A}}}{T_{\mathrm{A}}}$ because volume is kept constant.
$$\begin{gathered} Q=n{C}_{v}dT \\ T=\frac{Q}{n{C}_v} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{75}{300}=\frac{{\mathrm{P}}_{\mathrm{A}}\times 0.008\times 12.5}{5} \\ \Rightarrow P_{\mathrm{A}}=\frac{75\times 5}{300\times 0.008\times 12.5} \\ =12.5 \mathrm{cm} \mathrm{of} \mathrm{Hg} \\ \text{A}\mathrm{gain}, \frac{P}{T}\mathit{=}\frac{P_B}{T_B} [\mathrm{For} \mathrm{container} \mathrm{B}] \\ \Rightarrow \frac{75}{300}=\frac{P_{\mathrm{B}} \times 0.008\times 12.5}{10} \end{gathered}$$
P_B = 25 cm of Hg

The distance moved by the mercury,
P_B − P_A = 25 − 12.5 = 12.5 cm

Answer:

Given:
Mass of He, m_He = 0.1 g
γ₁ = 1.67
Molecular weight of He, M_He = 4 g/mol
M_H2 = ?
M_H2 = 2 g/mol
γ₂ = 1.4

Since it is an adiabatic environment and the system is not dong any external work, the amount of heat given will be used up entirely to raise its internal energy.
For He, dQ = dU = nC_vdT   ...(i)
$$\begin{gathered} =\frac{m}{4}\times \frac{\mathrm{R}}{\gamma -1}\times d\mathrm{T} \\ =\frac{0.1}{4}\times \frac{\mathrm{R}}{(1.67-1)}\times d\mathrm{T} \end{gathered}$$
For H₂, dQ =dU = nCvdT   ...(ii)
$$\begin{gathered} =\frac{m}{2}\times \frac{\mathrm{R}}{\gamma -1}\times d\mathrm{T} \\ =\frac{m}{2}\times \frac{\mathrm{R}}{1.4-1}\times d\mathrm{T}, \end{gathered}$$

where m is the required mass of H₂.

Since equal amount of heat is given to both gases; so dQ is same in both eq (i) and (ii), we get
$$\begin{gathered} \frac{0.1}{4}\times \frac{R}{0.67}dT \\ =\frac{m}{2}\times \frac{R}{0.4}\times dT \\ \Rightarrow m=\frac{0.1}{2}\times \frac{0.4}{0.67} \\ \Rightarrow m=0.0298\approx 0.03 \mathrm{g} \end{gathered}$$

Answer:

Initial pressure of the gas in both the vessels = P₀
Initial temperature of the gas in both the vessels = T₀
Initial volume = V₀
$\frac{\mathrm{C}p}{\mathrm{C}v}=\gamma$

(a) The temperature inside the diathermic vessel remains constant. Thus,
P₁V₁ = P₂V₂
⇒ P₀V₀ = P₂ × 2V₀
$\Rightarrow P_2=\frac{P_0}{2}$
Temperature = T₀
The temperature inside the adiabatic vessel does not remain constant.
So, for an adiabatic process,
T₁V₁^γ−1 = T₂V₂^γ−1
⇒ T₀V₀^γ−1 = T × (2V₀)^γ−1
⇒ T₂ = T₀ × 2^1−γ
⇒ P₁V₁^γ = P₂V₂^γ
P₀V₀^γ = P₂ × (2V₀)^γ
$\Rightarrow P_2=\left(\frac{P_0}{2^{\gamma }}\right)$

(b) When the valves are open, the temperature remains T₀ throughout, i.e. T₁ = T₂ = T₀ .

Also, pressure will be same throughout. Thus, P₁ = P_2. As, temperature has not changed on side 1, so pressure on this side will also not change (volume is also fixed due to fixed piston) and will be equal to $P_0$ (pressure is an intrinsic variable).
On side 2, pressure will change to accommodate the changes in temperature on this side.
So, P₀ = P₁ + P₂
⇒ 2P₁ = 2P₂
So, P₁ = P₂ = $\frac{{\mathrm{P}}_0}{2}$

Answer:

For an adiabatic process, PV^γ = Constant
So,  P₁V₁^γ = P₂V₂^γ   ...(i)
According to the problem,
V₁ + V₂ = V₀  ...(ii)
Using the relation in eq (ii) in eq (i), we get
P₁V₁^γ = P₂(V₀ − V₁)^γ
$\text{Or} {\left(\frac{P_1}{P_2}\right)}^{1/\gamma }=\frac{V_0-V_1}{V_1}$

$$\begin{gathered} V_1{P_{\mathit{1}}}^{\frac{\mathit{1}}{\gamma }}\mathit{=}{V}_0{P_{\mathit{2}}}^{\frac{\mathit{1}}{\gamma }}\mathit{-}{V}_1{P_2}^{\frac{\mathit{1}}{\gamma }} \\ {V}_1\left({P_1}^{\frac{\mathit{1}}{\gamma }}+{P_2}^{\frac{\mathit{1}}{\gamma }}\right)\mathit{=}{V}_0{P_2}^{\frac{\mathit{1}}{\gamma }} \\ \mathit{ }{V}_{\mathit{1}}\mathit{=}\frac{{P_2}^{{\displaystyle \frac{1}{\gamma }}}{V}_{\mathit{0}}}{{P_1}^{{\displaystyle \frac{1}{\gamma }}}+{P_2}^{{\displaystyle \frac{1}{\gamma }}}} \end{gathered}$$

Using equation (ii), we get
$V_2\mathit{=}\frac{{P_1}^{{\displaystyle \frac{1}{\gamma }}}{V}_{\mathit{0}}}{{P_1}^{{\displaystyle \frac{1}{\gamma }}}+{P_2}^{{\displaystyle \frac{1}{\gamma }}}}$

(b) Since the whole process takes place in adiabatic surroundings, the separator is adiabatic.
Hence, heat given to the gas in the left part is 0.

(c) There will be a common pressure 'P' when equilibrium is reached. The slid will move until the pressure on the two sides becomes equal.
P₁V₁^γ + P₂V₂^γ = PV₀^γ
For equilibrium, V₁ = V₂ = $\frac{{\mathrm{V}}_0}{2}$
Hence,
$$\begin{gathered} P_1{\left(\frac{V_{\mathit{0}}}{2}\right)}^{\gamma }+P_2{\left(\frac{V_{\mathit{0}}}{2}\right)}^{\gamma }\mathit{=}P\mathit{(}{V}_0{\mathit{)}}^{\gamma } \\ \text{Or} \mathit{ }P\mathit{=}{\left(\frac{{P_1}^{{\displaystyle \frac{1}{\gamma }}}+{P_2}^{{\displaystyle \frac{1}{\gamma }}}}{2}\right)}^{\mathrm{\gamma }} \end{gathered}$$

Answer:

Given:
Area of the tube, A = 1 cm² = 1 × 10⁻⁴ m²
Mass of the gas, M = 0.03 g = 0.03 × 10⁻³ kg
Initial pressure, P = 1 atm = 10⁵ Pascal
Initial length of the mercury column, L = 40 cm = 0.4 m
Final length of the mercury column, L₁ = 80 cm = 0.8 m
Final pressure, P' = 0.355 atm

The process is adiabatic. So,
P(V)^γ = P'(V')^γ
⇒ 1 × (A × 0.4)^γ = 0.355 × (A × 0.8)^γ
⇒ 1 × 1 = 0.355 × 2^γ 
$$\begin{gathered} \Rightarrow \frac{1}{0.355}=2^{\gamma } \\ \gamma \log 2=\log \left(\frac{1}{0.355}\right) \end{gathered}$$
⇒ γ = 1.4941

Speed of sound in the gas,
$$\begin{gathered} v=\sqrt{\frac{\gamma P}{\rho }}=\sqrt{\frac{\gamma P}{m\mathit{/}v}} \\ v=\sqrt{\frac{1.4941\times {10}^5}{m/v}}=\sqrt{\frac{1.4941\times {10}^5}{\left({\displaystyle \frac{0.03\times {10}^{-3}}{{10}^{-4}\times 1\times 0.4}}\right)}} \\ \Rightarrow v=446.33\approx 447 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

Answer:

Given:
Velocity of sound in hydrogen, V = 1280 m/s
Temperature, T = 0°C = 273 K

Density of H₂ = 0.089 kg/m³
R = 8.3 J/mol-K
At STP,
  P = 10⁵ Pa
We know:
$$\begin{gathered} V_{sound}\mathit{=}\sqrt{\frac{\gamma p}{\rho }} \\ 1280=\sqrt{\frac{\gamma \times {10}^5}{0.089}} \\ \text{Or} \gamma =\frac{1280\times 1280\times 0.089}{{10}^5} \\ =1.46 \\ \frac{C_p}{C_v}\mathit{=}\gamma \mathrm{or}\mathit{ }{C}_p\mathit{-}{C}_v\mathit{=}R \\ {C}_v\mathit{=}\frac{R}{\gamma \mathit{-}\mathit{1}}=\frac{8.3}{1.46-1} \\ =18.0 \mathrm{J}/\mathrm{mol}-\mathrm{K} \\ {C}_p\mathit{=}\gamma C_v=1.46\times 18.0 \\ =26.28 \approx 26.3/\mathrm{mol}-\mathrm{K} \end{gathered}$$

Answer:

Given:
Specific heat capacity at constant pressure, C_p = 5.0 cal/mol-K
C_p = 5.0 × 4.2 J/mol-K
C_p = 21 J/mol-K
Volume of helium, V = 22400 ${\mathrm{cm}}^3$ = 0.0224 ${\mathrm{m}}^3$
At STP, P = 1 atm = ${10}^5 \mathrm{Pa}$
The speed of sound in gas,
$$\begin{gathered} v\mathit{=}\sqrt{\frac{\mathit{\gamma }\mathit{p}}{\mathit{\rho }}}\mathit{=}\sqrt{\frac{\mathit{\gamma }\mathit{R}\mathit{T}}{\mathit{M}}}\mathit{=}\sqrt{\frac{\mathit{\gamma }\mathit{P}\mathit{V}}{\mathit{M}}} \\ {C}_{\mathit{p}}\mathit{=}\frac{\mathit{R}\mathit{\gamma }}{\mathit{\gamma }\mathit{-}\mathit{1}} \end{gathered}$$
Or 21(γ − 1) = 8.3γ
⇒ 21γ − 8.3γ = 21
⇒ 12.7γ = 21
$$\begin{gathered} \therefore \gamma =\frac{21}{12.7}=1.653 \\ v=\sqrt{\frac{1.653\times 1.0\times {10}^5\times 0.0224}{4\times {10}^{-3}}} \\ v=960 \mathrm{m}/\mathrm{s} \end{gathered}$$

Answer:

Given:
Density of the ideal gas, ρ = 1.7 × 10⁻³ g/cm³
                                    = 1.7 k/gm³
Pressure of the gas, P = 1.5 × 10⁵ Pa
R = 8.3 J/mol-K

Resonance frequency of the gas = 3.0 kHz
Node separation in the Kundt's tube
= $\frac{l}{2}$ = 6 cm
So, l = 2$\times$6 = 12 cm = 12 × 10⁻² m
So, V = fl = 3 × 10³ × 12 × 10⁻²
          = 360 m/s
Speed of sound, V = $\sqrt{\frac{\gamma p}{\rho }}$
Or $V^2=\frac{\gamma p}{\mathrm{\rho }}$
$$\begin{gathered} \therefore \mathit{ }\gamma \mathit{=}\frac{v^{\mathit{2}}\rho }{P}\mathit{=}\frac{(360{)}^2 \times 1.7\times {10}^{-3}}{1.5\times {10}^5} \\ =1.4688 \\ \mathrm{Using} C_p-C_v=R\mathit{ }and\mathit{ }\frac{C_p}{C_v}\mathit{=}\gamma \\ \mathrm{W}\text{e} \mathrm{know} \mathrm{that} \\ {C}_v\mathit{=}\frac{R}{\gamma \mathit{-}\mathit{1}}=\frac{8.3}{0.4688} \\ =17.7 \mathrm{J}/\mathrm{mol}-\mathrm{K} \\ {\mathrm{C}}_p=R\mathit{ }+ {\mathrm{C}}_v=8.3 + 17.7 = 26 \mathrm{J}/\mathrm{mol}-\mathrm{K} \end{gathered}$$