Question

Diagram shows the variation of internal energy $\left(\right.U\left.\right)$ with the pressure $\left(\right.P\left.\right)$ of $2.0 \, mole$ gas in cyclic process $abcda$ . The temperature of gas at $c$ and $d$ are $300 \, K$ and $500 \, K$ , respectively. Calculate the heat absorbed by the gas during the process.

• A. $440 \, R \, ln2$
• B. $400 \, R \, ln2$
• C. $430 \, R \, ln2$
• D. $414 \, R \, ln2$

Answer 1

Answer: (B)

Change in internal energy for cyclic process $\left(\Delta \text{U}\right) = 0$
For process $\text{a} \rightarrow \text{b} \text{, } \left(\text{P} - \text{constant}\right)$
$\text{W}_{\text{a} \rightarrow \text{b}} = \text{P} \Delta \text{V} = \text{n} \text{R} \Delta \text{T} = - 4 0 0 \text{ R}$
For process $\text{b} \rightarrow \text{c} \text{, } \left(\text{T} - \text{constant}\right)$
$\left(\text{W}\right)_{\text{b} \rightarrow \text{c}}=nRTln\left(\frac{P_{i}}{P_{f}}\right)=-2\text{R}\left(300\right)\text{ln}2$
For process $\text{c} \rightarrow \text{d} \text{, } \left(\text{P} - \text{constant}\right)$
$\text{W}_{\text{c} \rightarrow \text{d}}=\text{P}\Delta \text{V}=\text{nR}\Delta \text{T=}+400\text{R}$
For process $\text{d} \rightarrow \text{a} \text{, } \left(\text{T} - \text{constant}\right)$
$\left(\text{W}\right)_{\text{d} \rightarrow \text{a}}=nRTln\left(\frac{P_{i}}{P_{f}}\right)=+2\text{R}\left(5 0 0\right)\text{ln}2$
Net work,
$\left(\Delta \text{W}\right) = \left(\text{W}\right)_{\text{a} \rightarrow \text{b}} + \left(\text{W}\right)_{\text{b} + \text{c}} + \left(\text{W}\right)_{\text{c} \rightarrow \text{d}} + \left(\text{W}\right)_{\text{d} \rightarrow \text{a}}$
$\Delta \text{W} = 4 0 0 \text{ R ln 2}$
$\text{dQ} = \text{dU} + \text{dW}$ , first law of thermodynamics
$\text{dQ} = 4 0 0 \text{ R ln 2}$