Question

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

• A. $440Rln2$
• B. $400Rln2$
• C. $430Rln2$
• D. $414Rln2$

Answer 1

Answer: (B)

Change in internal energy for cyclic process $\left(\Delta \text{U}\right)=0$
For process $\text{a}\to \text{b}\text{, }\left(\text{P}-\text{constant}\right)$
${\text{W}}_{\text{a}\to \text{b}}=\text{P}\Delta \text{V}=\text{n}\text{R}\Delta \text{T}=-400\text{ R}$
For process $\text{b}\to \text{c}\text{, }\left(\text{T}-\text{constant}\right)$
${\left(\text{W}\right)}_{\text{b}\to \text{c}}=nRTln\left(\frac{P_i}{P_f}\right)=-2\text{R}\left(300\right)\text{ln}2$
For process $\text{c}\to \text{d}\text{, }\left(\text{P}-\text{constant}\right)$
${\text{W}}_{\text{c}\to \text{d}}=\text{P}\Delta \text{V}=\text{nR}\Delta \text{T=}+400\text{R}$
For process $\text{d}\to \text{a}\text{, }\left(\text{T}-\text{constant}\right)$
${\left(\text{W}\right)}_{\text{d}\to \text{a}}=nRTln\left(\frac{P_i}{P_f}\right)=+2\text{R}\left(500\right)\text{ln}2$
Net work,
$\left(\Delta \text{W}\right)={\left(\text{W}\right)}_{\text{a}\to \text{b}}+{\left(\text{W}\right)}_{\text{b}+\text{c}}+{\left(\text{W}\right)}_{\text{c}\to \text{d}}+{\left(\text{W}\right)}_{\text{d}\to \text{a}}$
$\Delta \text{W}=400\text{ R ln 2}$
$\text{dQ}=\text{dU}+\text{dW}$ , first law of thermodynamics
$\text{dQ}=400\text{ R ln 2}$