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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 $400 \mathrm{~K}, 700 \mathrm{~K}, 2500 \mathrm{~K}$ and $1100 \mathrm{~K}$ respectively. The work done by the gas during the process 1-2-3-4-1 is (universal gas constant is $R$ )
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Verified Answer
The correct answer is:
$1650 R$
We knows
$$
d Q=d u+d w
$$
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
$$
\begin{aligned}
& w_{2-3}=P_2\left(v_3-v_2\right) \\
& w_{4-1}=p_1\left(v_1-v_4\right)
\end{aligned}
$$
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
$$
\begin{aligned}
& =\frac{3 R}{2}(400+2500-700-1100) \\
& =\frac{3}{2} R(2900-1800) \\
& =\frac{3}{2} R(1100)=\frac{3300 R}{2} \\
& =1650 R
\end{aligned}
$$
$$
d Q=d u+d w
$$
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
$$
\begin{aligned}
& w_{2-3}=P_2\left(v_3-v_2\right) \\
& w_{4-1}=p_1\left(v_1-v_4\right)
\end{aligned}
$$
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
$$
\begin{aligned}
& =\frac{3 R}{2}(400+2500-700-1100) \\
& =\frac{3}{2} R(2900-1800) \\
& =\frac{3}{2} R(1100)=\frac{3300 R}{2} \\
& =1650 R
\end{aligned}
$$
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