An engine takes in 5 moles of air at 20ᵒC and 1 atm, and compresses it adiabatically to 1/10ᵗʰ of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be X kJ. The value of X to the nearest integer is________.

Solution

Physics sir · 4 steps

Use TV^(γ-1) = constant to get the final temperature, then Δ U = nC_VΔ T.

1. Set up the gas. A rigid diatomic molecule has 5 degrees of freedom, so C_V = 5/2R and γ = 7/5 = 1.4. The initial temperature in kelvin is T₁ = 20 + 273 = 293 K.

   T₁V₁^(γ-1) = T₂V₂^(γ-1) ⇒ T₂ = T₁(V₁/V₂)^(γ-1) = 293 × 10^(0.4)

2. Final temperature. 10^(0.4) ≈ 2.512, so T₂ ≈ 293 × 2.512 = 736 K. The rise in temperature is Δ T ≈ 736 - 293 = 443 K.
3. Internal energy. Δ U = nC_VΔ T = 5 × 5/2 × 8.314 × 443 ≈ 4.6 × 10⁴ J. It is positive because the work done on the gas during compression all goes into internal energy.
4. So. Δ U ≈ 46.04 kJ, so X rounds to 46.

The common slip: 3. Using T₁ = 20 in place of 293 K gives Δ T ≈ 30 K and a Δ U of about 3 kJ. The adiabatic relation needs absolute temperature.

The idea to keep. In an adiabatic process Q = 0, so Δ U = nC_VΔ T whatever the path. Find T₂ from TV^(γ-1) = constant, always with T in kelvin.

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