InterviewSolution
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InterviewSolution
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Calculate the fall in temperature of helium initially at `15^(@)C`, when it is suddenly expanded to `8 times` its original volume `(gamma=5//3)`. |
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Answer» Here, `T_(1)=273+15=288 K` , `T_(2)=?, V_(2)=8V_(1), gamma=5//3` As expansion is sudden //adiabatic `:. T_(2)V_(2)^(gamma-1)= T_(1)V_(1)^(gamma-1)` `:. T_(2)=T_(1)((V_(1))/(V_(2)))^(gamma-1) = 288((V_(1))/(8V_(1)))^((5/3-1))` or `logT_(2)= log288+2/3log(1/8)` `log288+2/3[log1-log8]` `logT_(2)= 2.4594+2/3[0-0.90031]` `1.8573` `T_(2)= antilog 1.8573= 71.99K` `:.` fall in temperature of helium `T_(1)-T_(2)=288-71.99= 216.01K` `=216.01^(@)C` `( :. "in magnitude", 1^(@)C=1K)` |
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