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Question: Two moles of an ideal monoatomic gas occupy a volume 2V at temperature 300 K, it expands to a volume...

Two moles of an ideal monoatomic gas occupy a volume 2V at temperature 300 K, it expands to a volume 4V adiabatically, then the final temperature of gas is

A

179 K

B

189 K

C

199 K

D

219 K

Answer

189 K

Explanation

Solution

: since in an adiabatic process

T1V1γ−1=T2V2γ−1T _ { 1 } V _ { 1 } ^ { \gamma - 1 } = T _ { 2 } V _ { 2 } ^ { \gamma - 1 }

T2=T1(V1V2)γ−1=300(2V4V)(53−1)T _ { 2 } = T _ { 1 } \left( \frac { V _ { 1 } } { V _ { 2 } } \right) ^ { \gamma - 1 } = 300 \left( \frac { 2 V } { 4 V } \right) ^ { \left( \frac { 5 } { 3 } - 1 \right) }

[∵V1=2V\left[ \because V _ { 1 } = 2 V \right. and V2=4VV _ { 2 } = 4 V for monoatomic gas γ=5/3]\left. \gamma = 5 / 3 \right]

=300(12)2/3=30022/3=188.98≈189 K= 300 \left( \frac { 1 } { 2 } \right) ^ { 2 / 3 } = \frac { 300 } { 2 ^ { 2 / 3 } } = 188.98 \approx 189 \mathrm {~K}