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

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

Step-by-step 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 [(V_(1))/(V_(2)) \right ]^(\gamma-1 ) = (T_(2))/(T_(1))`

`\left [10 \right ]^{(7)/(5)-1 } = (T_(2))/(300 K)`

`\left [10 \right ]^{(2)/(5) } = (T_(2))/(300 K)`

`2.512 = (T_(2))/(300 K)`

`T_(2) = 753.6 K`

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