A poly-atomic molecule (C_V=3 R, C_P=4 R, where R is gas constant) goes from phase space point A(P_A=10⁵ Pa, V_A=4 × 10⁻⁶ m³) to point B(P_B=5 × 10⁴ Pa, V_B=6 × 10⁻⁶ m³) to point C(P_C=10⁴ Pa, V_C=8 × 10⁻⁶ m³ ). A to B is an adiabatic path and B to C is an isothermal path.
The net heat absorbed per unit mole by the system is :

Solution

Physics sir · 4 steps

The adiabatic leg gives no heat at all, so the whole answer comes from the isothermal leg at 450 K.

1. Adiabatic A to B. On an adiabatic path no heat crosses the boundary, so Q_(AB) = 0. The gas cools from 500 K to 450 K using its own internal energy.
2. Isothermal B to C. The temperature stays at 450 K, so Δ U = 0 for an ideal gas. By the first law, the heat absorbed equals the work done, Q_(BC) = W_(BC).

   Q_(BC)/n = RT ln V_C/V_B = 450R ln(8 × 10⁻⁶)/(6 × 10⁻⁶) = 450R ln 4/3

3. Use the volumes. The stated pressures do not lie on one exact isotherm, so work with the volume ratio, as the options do. ln 4/3 = ln 4 - ln 3.
4. So. Net heat per mole = 0 + 450R(ln 4 - ln 3), which is option (b).

The idea to keep. For a multi-step process, find the heat leg by leg. Adiabatic legs give zero, and an isothermal leg gives Q = nRT ln(V_f/Vᵢ) at that leg's own temperature.

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