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(B) In the given $P-T$ diagram, process $AB$ is an isothermal process because the temperature $T$ is constant.For an ideal gas, $PV = nRT$. Since $T$

Vedclass · Accepted Answer

In process $AB$, heat is rejected out of the system.

Vedclass · Answer

(B) In the given $P-T$ diagram, process $AB$ is an isothermal process because the temperature $T$ is constant.For an ideal gas, $PV = nRT$. Since $T$ is constant, $PV = 	ext{constant}$, which implies $P \propto 1/V$.In process $AB$, pressure $P$ increases, so volume $V$ must decrease.Work done by the system $W = \int P \, dV$. Since $dV According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$.For an ideal gas, internal energy $U$ depends only on temperature. Since $T$ is constant in process $AB$, $\Delta U = 0$.Therefore, $\Delta Q = \Delta W$. Since $\Delta W Thus, option $B$ is correct.

An ideal gas is taken through the process shown in the figure. Which statement is true?

Similar Questions

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