Physics - Heat & Thermo Question with Solution | TestHub

Correct Answer: B

The process $A\to B$ is a straight line on the $P-V$ diagram.

From the graph:

Point A: $(V_A,P_A)=(V_0,3P_0)$

Point B: $(V_B,P_B)=(5V_0,6P_0)$

The equation of the line passing through A and B is $P=\frac{3P_0}{4V_0}V+\frac{9P_0}{4}$.

Work done ($W$) by the gas during the process is the area under the $P-V$ curve:

$$W={\int }_{V_A}^{V_B}PdV={\int }_{V_0}^{5V_0}\left(\frac{3P_0}{4V_0}V+\frac{9P_0}{4}\right)dV$$

$$W={\left[\frac{3P_0}{8V_0}{V}^2+\frac{9P_0}{4}V\right]}_{V_0}^{5V_0}$$

$$W=\left(\frac{3P_0}{8V_0}(25V_0^2)+\frac{9P_0}{4}(5V_0)\right)-\left(\frac{3P_0}{8V_0}(V_0^2)+\frac{9P_0}{4}(V_0)\right)$$

$$W=\left(\frac{75P_0{V}_0}{8}+\frac{45P_0{V}_0}{4}\right)-\left(\frac{3P_0{V}_0}{8}+\frac{9P_0{V}_0}{4}\right)$$

$$W=\left(\frac{75P_0{V}_0+90P_0{V}_0}{8}\right)-\left(\frac{3P_0{V}_0+18P_0{V}_0}{8}\right)$$

$$W=\frac{165P_0{V}_0}{8}-\frac{21P_0{V}_0}{8}=\frac{144P_0{V}_0}{8}=18P_0{V}_0$$

For a monoatomic ideal gas, the change in internal energy ($\Delta U$) is given by:

$$\Delta U=n{C}_V\Delta T$$

For 1 mole of monoatomic gas, $C_V=\frac{3}{2}R$.

Also, for an ideal gas, $PV=nRT$, so $\Delta T=\frac{\Delta (PV)}{nR}$.

Thus, $\Delta U=n\left(\frac{3}{2}R\right)\frac{(P_B{V}_B-P_A{V}_A)}{nR}=\frac{3}{2}(P_B{V}_B-P_A{V}_A)$.

$$P_A{V}_A=(3P_0)(V_0)=3P_0{V}_0$$

$$P_B{V}_B=(6P_0)(5V_0)=30P_0{V}_0$$

$$\Delta U=\frac{3}{2}(30P_0{V}_0-3P_0{V}_0)=\frac{3}{2}(27P_0{V}_0)=\frac{81}{2}{P}_0{V}_0$$

According to the First Law of Thermodynamics, the heat supplied ($Q$) is:

$$Q=W+\Delta U$$

$$Q=18P_0{V}_0+\frac{81}{2}{P}_0{V}_0=\frac{36P_0{V}_0+81P_0{V}_0}{2}=\frac{117}{2}{P}_0{V}_0$$

The specific heat ($C$) for this process is defined by $Q=nC\Delta T$.

Since $n=1$ mole, $Q=C\Delta T$.

We know $\Delta T=T_B-T_A=\frac{P_B{V}_B}{R}-\frac{P_A{V}_A}{R}=\frac{30P_0{V}_0}{R}-\frac{3P_0{V}_0}{R}=\frac{27P_0{V}_0}{R}$.

So,

$$C=\frac{Q}{\Delta T}=\frac{\frac{117}{2}{P}_0{V}_0}{\frac{27P_0{V}_0}{R}}$$

$$C=\frac{117}{2}\cdot \frac{R}{27}=\frac{117R}{54}$$

Dividing numerator and denominator by 9:

$$C=\frac{13R}{6}$$

The final answer is $\boxed{C=\frac{13R}{6}}$.