the volume of an ideal gas at temperature 27degree Celsius is changed to 1/4th of its initial value adiabatically. The new temperature of the gas is (gamma =1.5)
Answers
Answer:
600 k
Explanation:
T2/T1= (v1/V2) power gamma - 1
The new temperature of the gas if the final volume is changed to 1/4th of its initial volume adiabatically is 600 K or 327℃.
Explanation:
We know in an adiabatic process,
TV^(ϒ-1) = constant
Therefore, we can write,
T₁V₁^(ϒ-1) = T₂V₂^(ϒ-1)
⇒ T₁/T₂ = [V₂/V₁] ^(ϒ-1) ……. (i)
Where
T₁ = initial temperature = 27℃ = 27+273K = 300 K
V₁ = initial volume
V₂ = final volume = V₁/4 …. [given]
ϒ = 1.5
So, substituting the given values in eq. (i), we get
300/T₂ = [(V₁/4)/V₁]^(1.5-1)
⇒ 300/T₂ = [1/4]^(0.5)
⇒ 300/T₂ = √[0.25]
⇒ T₂ = 300/0.5
⇒ T₂ = 600 K or (600-273) = 327℃
Thus, the new temperature of the gas is 600 K or 327℃.
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