An ideal gas is taken through the process sho | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

For one complete cycle of a thermodynamic process on a gas as shown in the $P-V$ diagram,which of the following is correct?

What is the total work done in the given $P-V$ graph?

What is the net work done in the process shown in the $P-V$ diagram?

Which of the following graphs correctly represents the thermodynamic processes shown?

Vedclass Test Series

Exam Paper Generator

Online Exam Module