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Physics Question 47 – JEE-MAIN 2026

A vessel contains 0.15 m3 of a gas at pressure 8 bar and temperature 140 C with cp=3R and cv=2R. It is expanded adiabatically till pressure falls to 1 bar. The work done during this process is _______ k J. (R is gas constant)

Identify the type of thermodynamic process (adiabatic) and recall its defining characteristics and relevant formulas.

Step 1: Calculate Adiabatic Index (γ) and Final Temperature (T2)✦ Active

Given specific heats cp=3R and cv=2R. The adiabatic index is calculated as γ=cpcv. The initial temperature T1 must be converted to Kelvin.

γ=3R2R=1.5 T1=140 C+273=413 K T2=T1(P2P1)(γ1)/γ=413 K×(1 bar8 bar)(1.51)/1.5 T2=413×(18)0.5/1.5=413×(18)1/3=413×12=206.5 K
Step 2: Calculate the Number of Moles (n)○ Expand

Using the ideal gas law P1V1=nRT1, we can find the number of moles n. Ensure all units are in SI.

n=P1V1RT1=(8×105 Pa)×(0.15 m3)(8.314 J mol1 K1)×(413 K) n=1.2×1053433.98234.94 mol
Step 3: Calculate Work Done (W)○ Expand

The work done during an adiabatic process is given by the formula W=nR(T1T2)γ1. The final answer should be in kilojoules.

W=nR(T1T2)γ1=34.94 mol×8.314 J mol1 K1×(413 K206.5 K)1.51 W=34.94×8.314×206.50.5120000 J W=120 kJ
💡 Teacher's Secret Hint

Alternatively, one could calculate V2 using P1V1γ=P2V2γ and then use W=P1V1P2V2γ1.

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