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Question: The fall in temperature of helium gas initially at 20°C when it is suddenly expanded to 8 times its ...

The fall in temperature of helium gas initially at 20°C when it is suddenly expanded to 8 times its original volume is (γ=53)\left( \gamma = \frac { 5 } { 3 } \right)

A

70.25 K

B

71.25 K

C

72.25 K

D

73.25 K

Answer

73.25 K

Explanation

Solution

: since gas is suddenly expanded it means the process is adiabatic process, then

TiV1γ−1=T2V2γ−1T _ { i } V _ { 1 } ^ { \gamma - 1 } = T _ { 2 } V _ { 2 } ^ { \gamma - 1 }

Putting T1=273+20=293 K,V2=8V1T _ { 1 } = 273 + 20 = 293 \mathrm {~K} , V _ { 2 } = 8 V _ { 1 }

293=T28γ−1293 = T _ { 2 } 8 ^ { \gamma - 1 }

T2=2938γ−1=293853−1[∵γ=53]T _ { 2 } = \frac { 293 } { 8 ^ { \gamma - 1 } } = \frac { 293 } { 8 ^ { \frac { 5 } { 3 } - 1 } } \quad \left[ \because \gamma = \frac { 5 } { 3 } \right]

T2=29382/3=293(23)2/3=2934=73.25 KT _ { 2 } = \frac { 293 } { 8 ^ { 2 / 3 } } = \frac { 293 } { \left( 2 ^ { 3 } \right) ^ { 2 / 3 } } = \frac { 293 } { 4 } = 73.25 \mathrm {~K}