1 mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27° C. The work done is 3 kJ mol⁻¹. The final temperature of the gas is ________ K (Nearest integer).
Given C_V=20 J mol⁻¹ K⁻¹

Solution

Chemistry ma’am · 4 steps

No heat flows in an adiabatic change, so all the work the gas does comes out of its internal energy.

1. First law. For an adiabatic process q = 0, so Δ U = q + w = w. The gas expands, which means it does work on the surroundings. So w = -3 kJ mol⁻¹ = -3000 J mol⁻¹.
2. Link to temperature. For an ideal gas, internal energy depends only on temperature: Δ U = nC_VΔ T. This holds for any process, adiabatic included.

   nC_V(T₂ - T₁) = w ⇒ 1 × 20 × (T₂ - 300) = -3000

3. Solve. T₁ = 27 + 273 = 300 K. Then T₂ - 300 = -150, so T₂ = 150 K.
4. So. The gas cools as it pays for the work from its own internal energy. The final temperature is 150 K.

The common slip: 450. Treating the work as positive (energy gained) makes the gas heat up and gives 300 + 150 = 450 K. Expansion work is done by the gas, so Δ U is negative.

The idea to keep. Adiabatic means q = 0, so w = Δ U = nC_VΔ T. An expanding gas does work, loses internal energy and cools. Always convert °C to kelvin first.

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