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A diatomic gas initially at 18°C is compressed adiabatically to one eighth of its original volume. The temperature after compression will be (a) 18° C (b) 668.4° K (c) 395.4° C (d) 144° C



A diatomic gas initially at 18°C is compressed adiabatically to one eighth of its original volume. The temperature after compression will be


        (a)    18° C


        (b)    668.4° K


        (c)    395.4° C


        (d)    144° C


Grade:10

2 Answers

Simran Bhatia
348 Points
7 years ago

(b)

Initial temperature (T1) = 18°C = 291 K

Let Initial volume (V1) = V

Final volume (V2) =  

According to adiabatic process,

TVg-1 = constant

According to question, T1 V1y-1 = T2 V2y-1

Rishi Sharma
askIITians Faculty 646 Points
10 months ago
Dear Student,
Please find below the solution to your problem.

T1 ​= 18oC = (273+18) = 291K
and V2​ = V1​/8
We know that TV^(γ−1) = constant
or, T2​V2^(γ−1) ​= T1​V1^(γ−1​)
T2 ​= T1​(V2/​V1​​)^(γ−1)
= 291 × (8)^(1.4−1)
= 668.5K
= 395.5oC

Thanks and Regards

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