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 $400\, K ,\, 700\, K ,\, 2500\, K$ and $1100\, K$ 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
$d Q=du + dw$
and we also known $d u=0$ for cyclio process so that
$d Q=d w$
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 $p V=n R T=\frac{3 \times T}{2}$
Hence, total work done
$=\frac{3 R}{2}(400+2500-700-1100)$
$=\frac{3}{2} R(2900-1800)$
$=\frac{3}{2} R(1100)=\frac{3300 R}{2}$
$=1650\, R$