The temperature of 3.00 mol of an ideal diatomic gas is increased by 40.0°C without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is x/10. Then the value of x (round off to the nearest integer) is ___________. (Given R = 8.31 J mol⁻¹ K⁻¹)

Solution

Physics sir · 4 steps

At constant pressure, Δ U and W share the same nRΔ T, so their ratio is just C_V/R.

1. Count degrees of freedom. A diatomic molecule that moves and rotates but does not vibrate has 3 translational and 2 rotational degrees of freedom, so 5 in all. That gives C_V = 5/2R.
2. Write both energies. Internal energy change: Δ U = nC_VΔ T = 5/2nRΔ T. Work at constant pressure: W = PΔ V = nRΔ T.

   (Δ U)/W = (5/2nRΔ T)/(nRΔ T) = 5/2 = 25/10

3. Check with numbers. Δ U = 3 × 2.5 × 8.31 × 40 = 2493 J and W = 3 × 8.31 × 40 = 997.2 J. Their ratio is 2.5. A rise of 40 °C is also a rise of 40 K.
4. So. x/10 = 25/10, so x = 25.

The common slip: 35. Using C_P = 7/2R in place of C_V for Δ U gives a ratio of 7/2, so x = 35. That ratio is Q/W, not Δ U/W.

The idea to keep. For an isobaric process, Δ U : W : Q = C_V : R : C_P. Get C_V from the degrees of freedom and the ratio follows without any arithmetic.

Answer from the NTA final answer key. Solution by PrepWiser.