The pressure P₁ and density d₁ of diatomic gas (γ=7/5) changes suddenly to P₂(>P₁) and d₂ respectively during an adiabatic process. The temperature of the gas increases and becomes ________ times of its initial temperature. (given d₂/d₁=32)

Solution

Physics sir · 4 steps

Density is just the inverse of volume, so turn TV^(γ-1) into a relation between T and d.

1. Volume from density. For a fixed mass of gas, d = m/V, so V ∝ 1/d. A sudden change means no time for heat flow, so the process is adiabatic.
2. Adiabatic relation. For an adiabatic process TV^(γ-1) is constant. Put V ∝ 1/d into it and you get T ∝ d^(γ-1).

   T₂/T₁ = (d₂/d₁)^(γ-1) = (32)^(2/5) = (2⁵)^(2/5) = 2²

3. Cross-check. From PV^γ constant, P₂/P₁ = 32^(7/5) = 2⁷ = 128. Ideal gas gives T ∝ P/d, so T₂/T₁ = 128/32 = 4, which agrees.
4. So. The temperature becomes 4 times its initial value.

The common slip: 128. Using the exponent γ instead of γ - 1 gives 32^(7/5) = 128. That is the pressure ratio, not the temperature ratio.

The idea to keep. In adiabatic questions with density, replace V by 1/d: then T ∝ d^(γ-1) and P ∝ d^γ. Writing 32 as 2⁵ keeps the powers easy.

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