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At 27°c ,a gas is compressed suddenly such that its pressure becomes 1/8th of its original pressure. Final temperature will be if gamma =5/3


2 years ago

## Answers : (2)

Arun
24738 Points
							Dear studentIn an adiabatic processPV$\gamma$ = constantP(nRT/P)$\gamma$ = constantP1-$\gamma$ T$\gamma$ = constantNow(P1/P2)1-$\gamma$/$\gamma$ = T2/T1Put the value and you will get the answer. RegardsArun (askIITians forum expert)

2 years ago
ankit singh
596 Points
							T^{\gamma}P^{1-\gamma} = constantT_{1}^{\gamma}P_{1}^{1-\gamma}=T_{2}^{\gamma}P_{2}^{1-\gamma}\dfrac{P_{1}^{1-\gamma}}{P_{2}^{1-\gamma}}=\dfrac{T_{2}^{\gamma}}{T_{1}^{\gamma}}(\dfrac{P_{1}}{P_{2}})^{1-\gamma}=(\dfrac{T_{2}}{T_{1}})^{\gamma}(\dfrac{P\times8}{P})^{1-\dfrac{5}{3}}=(\dfrac{T_{2}}{300})^{\dfrac{5}{3}}T_{2}= 300\times((8)^{-\dfrac{2}{3}})^{\dfrac{3}{5}}T_{2}= 130.58\ KT_{2}= 130.58-273 = -142.42\ CT_{2}=-142\ C

2 months ago
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