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  • An ideal gas is taken through a cyclic thermodynamic process involving four steps. The amounts of heat involved in these steps are <img alt="\mathrm{Q_1=5960 \mathrm{~J}, Q_2=-5585 \mathrm{~J}, Q_3

An ideal gas is taken through a cyclic thermodynamic process involving four steps. The amounts of heat involved in these steps are \mathrm{Q_1=5960 \mathrm{~J}, Q_2=-5585 \mathrm{~J}, Q_3=-2980 \mathrm{~J} \text {, and } Q_4=3645 \mathrm{~J}} respectively. The corresponding amounts of work done are \mathrm{W_1=2200 \mathrm{~J}, W_2=-825 \mathrm{~J} \text { and } W_3=-1100 \mathrm{~J} \text {, and } W_4} respectively. The efficiency of the cycle is η. Then

Option: 1

\mathrm{W_4=765 \mathrm{~J}}


Option: 2

\mathrm{W_4=275 \mathrm{~J}}


Option: 3

\eta \simeq 11 \%


Option: 4

Both a and c


Answers (1)

best_answer

Since\mathrm{W_2 \text { and } W_3} are negative, it means that the work is done on the gas. Hence \mathrm{Q_2 \text { and } Q_3} are negative which implies that heat is evolved in processes 2 and 3. Since\mathrm{Q_1 \text { and } Q_4} are positive, heat I absorbed by the gas in processes 1 and 4. As \mathrm{\(Q_1+Q_4)} is greater than\mathrm{\(Q_21+Q_3)} the gas absorbs a net amount of heat energy in a complete cycle, which is given by

\mathrm{\begin{aligned} & \Delta Q=Q_1+Q_2+Q_3+Q_4 \\ & =5960-5585-2980+3645 \\ & =1040 \text { joule } \end{aligned}}

The net work done by the gas is

\mathrm{\begin{aligned} & \Delta W=W_1+W_2+W_3+W_4 \\ & =2200-825-1100+W_4 \\ & =\left(275+W_4\right) \text { joule } \end{aligned}}

Since the process is cyclic, the change in internal energy ?U = 0. From the first law of
thermodynamics,
We have

\mathrm{\begin{aligned} & \Delta W=\Delta Q-\Delta U=\Delta Q \\ & \text { Or } 275+W_4=1040 \text { or } W_4=1040-275 \\ & =765 \mathrm{I} \end{aligned}}

Efficiency of the cycle is defined as

\mathrm{h=\frac{\text { net work done by the gas }}{\text { total heat absorbed by the gas }}}

\mathrm{\begin{aligned} & =\frac{\Delta W}{Q_1+Q_4}=\frac{275+765}{5960+3645} \\ & =\frac{1040}{9605}=0.1083=10.83 \% \\ & \simeq 11 \% \end{aligned}}

Posted by

Ritika Harsh

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