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:
[tex]V^{\gamma-1 }T = constant[/tex]
[tex]V_{1}^{\gamma-1 }T_{1} = V_{2}^{\gamma-1 }T_{2}[/tex]
[tex]\left [\frac{V_{1}}{V_{2}} \right ]^{\gamma-1 } = \frac{T_{2}}{T_{1}}[/tex]
[tex]\left [10 \right ]^{\frac{7}{5}-1 } = \frac{T_{2}}{300 K}[/tex]
[tex]\left [10 \right ]^{\frac{2}{5} } = \frac{T_{2}}{300 K}[/tex]
[tex]2.512 = \frac{T_{2}}{300 K}[/tex]
[tex]T_{2} = 753.6 K[/tex]
Therefore, the final temperature of the given ideal diatomic gas: T₂ = 753.6 K
