Question

Three moles of an ideal monoatomic gas undergoes a cyclic process as shown in the figure. The temperature of the gas in different states marked as $1,2,3$ and 4 are $400K,700K,2500K$ and $1100K$ respectively. The work done by the gas during the process $1-2-3-4-1$ is (universal gas constant is $R$)

• A. 1650 R
• B. 550 R
• C. 1100 R
• D. 2200 R

Answer 1

Answer: (A)

We knows
$dQ=du+dw$
and we also known $du=0$ for cyclio process so that
$dQ=dw$
Here, in given condition the work done during is a basic process
$w_{2-3}=P_2\left(v_3-v_2\right)$
$w_{4-1}=p_1\left(v_1-v_4\right)$
Total work done $=p_2\left(v_3-v_2\right)+p_1\left(v_1-v_4\right)$
From gas equation $pV=nRT=\frac{3\times T}{2}$
Hence, total work done
$=\frac{3R}{2}(400+2500-700-1100)$
$=\frac{3}{2}R(2900-1800)$
$=\frac{3}{2}R(1100)=\frac{3300R}{2}$
$=1650R$