Question

One mole of a monatomic ideal gas is expanded by a process described by \( PV^3 = C \), where \( C \) is a constant. The heat capacity of the gas during the process is given by (R is the gas constant)

• A. 2R
• B. 2.5R
• C. 1.5R
• D. R

Hint:

For a monatomic ideal gas, the heat capacity during a polytropic process can be derived using the equation \( C = \frac{3R}{2} \), considering the value of \( \gamma \).

Answer 1

The Correct Option is D

Step 1: For a polytropic process \( PV^{n} = C \), the molar heat capacity can be written as \( C = \dfrac{R(\gamma-n)}{(\gamma-1)(1-n)} \), where \( \gamma \) is the adiabatic index.
Step 2: For a monatomic gas, \( \gamma = \dfrac{5}{3} \), and here \( n = 3 \), so \( \gamma - n = \dfrac{5}{3}-3 = -\dfrac{4}{3} \), and \( (\gamma-1)(1-n) = \dfrac{2}{3}\times(-2) = -\dfrac{4}{3} \).
Step 3: Dividing, \( C = R\left(-\dfrac{4}{3}\right)\Big/\left(-\dfrac{4}{3}\right) = R \).
\[ \boxed{C = R} \]

Answer 2

We can derive the heat capacity directly from the first law of thermodynamics instead of using a ready-made formula. For one mole of gas, \( dQ = C_v\,dT + P\,dV \). From \( PV^{n} = \text{constant} \), differentiating gives \( V\,dP + nP\,dV = 0 \), so \( dP = -\dfrac{nP}{V}dV \). Using the ideal gas law \( PV = RT \), differentiating gives \( P\,dV + V\,dP = R\,dT \). Substituting \( dP \) from above,

\[ P\,dV - nP\,dV = R\,dT \implies P\,dV = \frac{R\,dT}{1-n} \]

So the heat added is \( dQ = C_v\,dT + \dfrac{R\,dT}{1-n} \), giving \( C = C_v + \dfrac{R}{1-n} \). Let's check each option using this relation with \( C_v = \dfrac{3}{2}R \) and \( n=3 \).

1. 2R: This would require \( \dfrac{R}{1-n} = \dfrac{1}{2}R \), that is \( 1-n=2 \), meaning \( n=-1 \), not the given \( n=3 \). So this option is incorrect.
2. 2.5R: This would require \( \dfrac{R}{1-n}=R \), meaning \( n=0 \) (an isobaric process), which is not the case here. So this option is incorrect.
3. 1.5R: This equals \( C_v \) alone, which would only be the heat capacity if the process added no extra work term, that is, if \( \dfrac{R}{1-n}=0 \), which never happens for finite \( n \). So this option is incorrect.
4. R: Substituting \( C_v=\dfrac{3}{2}R \) and \( n=3 \) directly gives \( C = \dfrac{3}{2}R + \dfrac{R}{1-3} = \dfrac{3}{2}R-\dfrac{1}{2}R = R \), which matches exactly.

Therefore, the correct answer is R.