Question

One mole of a monoatomic ideal gas undergoes the process $A \to B$ as shown in the given $P$–$V$ diagram. The specific heat capacity in the process is

• A. $\dfrac{13R}{3}$
• B. $\dfrac{13R}{6}$
• C. $\dfrac{7R}{3}$
• D. $\dfrac{2R}{3}$

Hint:

For any thermodynamic process, the specific heat is found from \[ C=\dfrac{\Delta Q}{\Delta T} \] using $\Delta Q=\Delta U+W$.

Answer 1

The Correct Option is B

Step 1: Read the coordinates of points $A$ and $B$ from the $P$–$V$ diagram: \[ A(V_0,\,3P_0), \qquad B(5V_0,\,6P_0) \] For one mole of ideal gas, \[ T=\frac{PV}{R} \]
Step 2: Temperatures at $A$ and $B$: \[ T_A=\frac{3P_0V_0}{R}, \qquad T_B=\frac{6P_0\cdot 5V_0}{R}=\frac{30P_0V_0}{R} \] \[ \Delta T=T_B-T_A=\frac{27P_0V_0}{R} \]
Step 3: Work done in the process $A\to B$. Since the path is a straight line, \[ W=\text{average pressure}\times \Delta V \] \[ W=\frac{3P_0+6P_0}{2}\,(5V_0-V_0) =\frac{9P_0}{2}\cdot 4V_0 =18P_0V_0 \]
Step 4: Change in internal energy for a monoatomic ideal gas: \[ \Delta U=\frac{3}{2}R\Delta T =\frac{3}{2}R\cdot\frac{27P_0V_0}{R} =\frac{81}{2}P_0V_0 \]
Step 5: Heat supplied using the first law of thermodynamics: \[ \Delta Q=\Delta U+W =\frac{81}{2}P_0V_0+18P_0V_0 =\frac{117}{2}P_0V_0 \]
Step 6: Specific heat capacity of the process: \[ C=\frac{\Delta Q}{\Delta T} =\frac{\frac{117}{2}P_0V_0}{\frac{27P_0V_0}{R}} =\frac{117}{54}R =\frac{13R}{6} \]