Q138 / 160 · TS EAMCET 2013 · Physics

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 $R$ ) Figure

  • Correct answer
    A
    $1650 R$
  • B
    $550 R$
  • C
    $1100 R$
  • D
    $2200 R$

Solution

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 $$ \begin{aligned} & =\frac{3 R}{2}(400+2500-700-1100) \\ & =\frac{3}{2} R(2900-1800) \\ & =\frac{3}{2} R(1100)=\frac{3300 R}{2} \\ & =1650 R \end{aligned} $$

Hence, the correct answer is option: 'A'.

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