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JEE MAIN 2026 · Thermodynamics

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 35

Updated
Single correct+4 / -1Answer matches NTA’s final key

One mole of diatomic gas having rotational modes only is kept in a cylinder with a piston system. The cross-section area of the cylinder is 4 cm². The gas is heated slowly to raise the temperature by 1.2 °C during which the piston moves by 25 mm. The amount of heat supplied to the gas is ______ J. (Atmospheric pressure = 100 kPa, RR = 8.3 J/mol. K) (Neglect mass of the piston)

Choose one

Solution

Solution

NTA accepted two options here. The final answer key credits both 24.8 and 25, so either earns the marks.

The working the question points to:

  • "Rotational modes only" means no vibration: 3 translational + 2 rotational degrees of freedom, so CV=52RC_V = \frac52 R.
  • Change in internal energy: ΔU=nCVΔT=1×52×8.3×1.2=24.9\Delta U = nC_V\Delta T = 1 \times \frac52 \times 8.3 \times 1.2 = 24.9 J.
  • Work against the atmosphere (massless piston): W=P ΔV=105×(4×10−4×0.025)=1W = P\,\Delta V = 10^5 \times (4 \times 10^{-4} \times 0.025) = 1 J.

By the first law Q=ΔU+W≈25.9Q = \Delta U + W \approx 25.9 J, which matches neither listed value exactly. Both accepted values sit within rounding of ΔU≈24.9\Delta U \approx 24.9 J, as if the work term had been left out. NTA has not published its reasoning; the official key is what settles the marks.

Answer: 24.8 or 25 (both accepted).

Answer checked against the official answer key. The solution is our own working.

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