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 $4\ \text{cm}^2$. The gas is heated slowly to raise the temperature by $1.2^\circ\text{C}$ during which the piston moves by $25$ mm. The amount of heat supplied to the gas is ______ J.
(Atmospheric pressure $= 100\ \text{kPa}$, $R = 8.3\ \text{J/mol.K}$) (Neglect mass of the piston)
Answer: (B) $10.96$
Concept. First law $Q = \Delta U + W$; the massless piston means the gas pushes against atmospheric pressure.
Step 1. Work. $W = pA\Delta x = 10^5 \times 4 \times 10^{-4} \times 0.025 = 1.0$ J.
Step 2. Internal energy. The official answer counts only the two rotational modes ("rotational modes only"), $f = 2$:
$$\Delta U = n\frac{f}{2}R\Delta T = 8.3 \times 1.2 = 9.96\ \text{J}$$
Step 3. Heat. $Q = 9.96 + 1.0 = 10.96$ J.
(With the usual 5 degrees of freedom, $\Delta U = 24.9$ J, which matches no option. The data are also not fully consistent: at constant pressure $p\Delta V$ should equal $nR\Delta T = 9.96$ J.)
Answer. $10.96$ (Option B).
Solution by Sreeraj P, M.Sc Physics
