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An ideal gas at 27 0C is compressed adiabatically to 8/27 of its original volume. if =5/3, then the rise in temperature isa. 450 0C b. 375 0Cc. 225 0Cd. 405 0C

Jitender Pal , 10 Years ago
Grade 9
anser 3 Answers
Abhishek

Last Activity: 7 Years ago

375 C ixffg*+₹₹66₹(₹+_-"-₹7_&-₹7647)_+_7)4)_7425#6₹7)_()34217_867(₹4##;68₹6₹+"(₹-"++6_+&9-)&8587252&@-?₹+(-(`_74(`(₹5@38/_+₹??"-

Hamza

Last Activity: 7 Years ago

Since it is a adiabaric processT×V^(gamma-1)Gamma for a monoatomic gas=5/3T°V°^(2/3)=T`(8V°/27)^(2/3)So T`=675 Kand change in temp=(675-300)=375K

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the solution to your problem.
 
Since it is a adiabaric process
Hence, TVγ-1 = constant
Where, γ = 5/3
Now,
T°V°2/3 = T’(8V°/27)2/3
So T’ = 675 K
Hence, change in temp = (675 – 300) = 375K
Hence, option b) is correct.
 
Thanks and regards,
Kushagra

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