Question

$100moles$ of an ideal monatomic gas undergoes the thermodynamic process as shown in the figure
A $\to$ B : isothermal expansion
B $\to$ C : adiabatic expansion
C $\to$ D : isobaric compression
D $\to$ A : isochoric process
The heat transfer along the process AB is $9\times {10}^{4}J$ . The net work done by the gas during the cycle is [Take $R=J{K}^{-1}mo{l}^{-1}$]

• A. $-0.5\times {10}^{4}J$
• B. $+0.5\times {10}^{4}J$
• C. $-5\times {10}^{4}J$
• D. $+5\times {10}^{4}J$

Answer 1

The correct option is D $+5\times {10}^{4}J$

Given : $W_{AB}=9\times {10}^{4}J$

Also $n=100\gamma =\frac{5}{3}$ (monoatomic)

$W_{BC}=\frac{P_2{V}_2-P_1{V}_1}{1-\gamma }=\frac{2\left({10}^5\right)-2.4\times {10}^5}{1-\frac{5}{3}}$

$W_{BC}=6\times {10}^{4}J$

$W_{CD}=-1\times {10}^{5}J$

$W_{DA}=0$ (as $\Delta V=0$)

Total work done , $W=9\times {10}^4+6\times {10}^4-1\times {10}^{5}J=+5\times {10}^{4}J$