Starting at temperature 300 K, one mole of an
ideal diatomic gas (γ = 1.4) is first compressed
adiabatically from volume V₁ to V₂ = V₁/16. It is
then allowed to expand isobarically to volume 2V₂. If all the processes are the quasi-static then
the final temperature of the gas (in ᵒK) is (to the nearest integer) _____.

Solution

Physics sir · 4 steps

Adiabatic step: use TV^(γ-1) constant. Then at constant pressure, doubling the volume doubles T.

1. Adiabatic compression. For a quasi-static adiabatic process on an ideal gas, TV^(γ-1) stays constant. Here γ - 1 = 0.4.

   T₂ = T₁(V₁/V₂)^(γ-1) = 300 × 16^(0.4)

2. Put in numbers. 16^(0.4) = 2^(1.6) = 2 × 2^(0.6) ≈ 2 × 1.5157 = 3.0314. So T₂ = 300 × 3.0314 ≈ 909.4 K.
3. Isobaric expansion. At constant pressure, V ∝ T for an ideal gas. The volume goes from V₂ to 2V₂, so the temperature doubles: T₃ = 2 × 909.4 ≈ 1818.9 K.
4. So. The final temperature is about 1818.9 K, which rounds to 1819.

The common slip: 909. Stopping after the adiabatic compression gives 909, the temperature at V₂. The isobaric expansion to 2V₂ still doubles it.

The idea to keep. Pick the adiabatic relation that links the two quantities you know: TV^(γ-1) for temperature and volume. For any isobaric step, T scales directly with V.

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