Question

An ideal diatomic gas starting at room temperature T1 = 300 K and atmospheric pressure p1 = 1.0 atm is compressed adiabatically to 1/10 of its original volume. What is the final temperature of the gas?

Answer 1

Answer:

The final temperature of the given ideal diatomic gas: T₂ = 753.6 K

Explanation:

Given: Atmospheric pressure: P = 1.0 atm

Initial Volume: V₁ , Final Volume: V₂ = V₁ (1/10)

⇒ V₁ / V₂ = 10

Initial Temperature: T₁ = 300 K, Final temperature: T₂ = ? K

 

For a diatomic ideal gas: γ =  7/5

For an adiabatic process:

$V^{\gamma-1 }T = constant$

$V_{1}^{\gamma-1 }T_{1} = V_{2}^{\gamma-1 }T_{2}$

$\left [\frac{V_{1}}{V_{2}} \right ]^{\gamma-1 } = \frac{T_{2}}{T_{1}}$

$\left [10 \right ]^{\frac{7}{5}-1 } = \frac{T_{2}}{300 K}$

$\left [10 \right ]^{\frac{2}{5} } = \frac{T_{2}}{300 K}$

$2.512 = \frac{T_{2}}{300 K}$

$T_{2} = 753.6 K$

Therefore, the final temperature of the given ideal diatomic gas: T₂ = 753.6 K