Question
GeneralGeneralGeneral

One mole diatomic gas at 1.23 ATM is compressed adiabatically from 40L to 20L. What is the new pressure and new temperature?

Verified Answer

For an adiabatic process, we use the relations:

  • P₁V₁^γ = P₂V₂^γ
  • T₁V₁^(γ−1) = T₂V₂^(γ−1)

For a diatomic gas, γ (gamma) = 7/5 = 1.4

Given:

  • P₁ = 1.23 atm
  • V₁ = 40 L
  • V₂ = 20 L
  • γ = 1.4

Step 1: Find new pressure (P₂)

P₂ = P₁ × (V₁ / V₂)^γ

P₂ = 1.23 × (40 / 20)^1.4
P₂ = 1.23 × (2)^1.4

2^1.4 ≈ 2.64

P₂ ≈ 1.23 × 2.64 ≈ 3.25 atm

Step 2: Find new temperature (T₂)

T₂ / T₁ = (V₁ / V₂)^(γ−1)

T₂ / T₁ = (2)^(0.4)

2^0.4 ≈ 1.32

So,

T₂ ≈ 1.32 × T₁

Final Answer:

  • New Pressure ≈ 3.25 atm
  • New Temperature ≈ 1.32 × initial temperature

Key Takeaway:

In adiabatic compression, pressure and temperature both increase, and γ (heat capacity ratio) plays a crucial role.