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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2025 · 3 April · Evening Shift · Physics · Thermodynamics

An ideal gas exists in a state with pressure P_0 volume V_0 .It is...

An ideal gas exists in a state with pressure P0P_0 volume V0\mathrm{V}_0.It is isothermally expanded to 4 times of its initial volume \left(\mathrm{V}_0\right)$, then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is :An ideal gas exists in a state with pressure P_0 volume V_0 .It is isothermally expanded to 4 times of its initial

Options

  1. AP0 V0(2ln⁡2−0.75)\mathrm{P}_0 \mathrm{~V}_0(2 \ln 2-0.75)
  2. BP0 V0(ln⁡2−0.75)\mathrm{P}_0 \mathrm{~V}_0(\ln 2-0.75)
  3. CP0 V0(ln⁡2−0.25)\mathrm{P}_0 \mathrm{~V}_0(\ln 2-0.25)
  4. DP0V0(2ln⁡2−0.25)P_0 V_0(2 \ln 2-0.25)

Correct answer

A. P0 V0(2ln⁡2−0.75)\mathrm{P}_0 \mathrm{~V}_0(2 \ln 2-0.75)

Step-by-step solution

An ideal gas exists in a state with pressure P_0 volume V_0 .It is isothermally expanded to 4 times of its initial
ω1=P0v0ln⁡4ω2=P04(−3v0)=−3P0v04\begin{aligned} & \omega_1=P_0 v_0 \ln 4 \\ & \omega_2=\frac{P_0}{4}\left(-3 v_0\right)=-\frac{3 P_0 v_0}{4}\end{aligned}
ω3=0QT=ΔUcyclic +ω\begin{aligned} & \omega_3=0 \\ & \mathrm{Q}_{\mathrm{T}}=\Delta \mathrm{U}_{\text{cyclic }}+\omega\end{aligned}
QT=ω(ΔUcyclic =0)\mathrm{Q}_{\mathrm{T}}=\omega \quad\left(\Delta \mathrm{U}_{\text{cyclic }}=0\right)
QT=P0v0(ln⁡4−34)=P0v0(2ln⁡2−0.75)\begin{aligned} & \mathrm{Q}_{\mathrm{T}}=\mathrm{P}_0 \mathrm{v}_0\left(\ln 4-\frac{3}{4}\right) \\ & =\mathrm{P}_0 \mathrm{v}_0(2 \ln 2-0.75)\end{aligned}
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