Question

0.2 moles of an ideal gas is taken round the cycle ABC as shown in the figure. The path B → C is an adiabatic process, A → B is an isochoric process and C → A is an isobaric process. The temerature at A and B are T = 300 K and T = 500 K and pressure at A is 1 atm and volume at A is 4.9 L. The volume at C is (Given : $\gamma=\frac{Cp}{C_{V}}=\frac{5}{3}, R = 8.205 × 10^{-2}$ L atm mol$^{-1}$ $K^{-1}, \left(\frac{3}{2}\right)^{^{2/5}} =0.81$

• A. 6.9 L
• B. 6.6 L
• C. 5.5 L
• D. 5.8 L

Answer 1

The Correct Option is B

For A $→$ B, Isochoric process $\frac{P_{A}}{T_{A}}=\frac{P_{B}}{T_{B}}$ $P_{B}=\frac{T_{B}}{T_{A}} P_{A}$ $P_{B}=\frac{500}{300}\times1=\frac{5}{3}\, atm \quad\quad\quad\quad\quad ...\left(i\right)$ For B $\rightarrow$ C, Adiabatic process $\therefore \frac{T^{\gamma}_{C}}{P_{C}^{\gamma-1}}=\frac{T^{\gamma}_{B}}{P_{B}^{\gamma-1}}$ $T_{C}=\left(\frac{P_{C}}{P_{B}}\right)^{^{\frac{\gamma-1}{\gamma}}}\times T_{B}$ $=\left[\frac{1}{5/3}\right]^{^{\frac{\left(5/3\right)-1}{\left(5/3\right)}}}\times500 \quad\quad\quad\quad\quad\left(Usinig \left(i\right)\right)$ $=\left(\frac{3}{5}\right)^{2/5} \times500 \quad\quad\quad\quad\quad ...\left(ii\right)$ For C $\to$ A, Isobaric process $\therefore \frac{V_{C}}{T_{C}}=\frac{V_{A}}{T_{A}}$ $V_{C}=V_{A}\times\frac{T_{C}}{T_{A}}=4.9\times\left(\frac{3}{5}\right)^{2/5}\times500\times\frac{1}{300} \quad \left(Using \left(ii\right)\right)$ $V_{c}=4.9\times0.81\times\frac{5}{3}=6.6 L$