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Question: One mole of an ideal monoatomic gas at temperature ![](https://cdn.pureessence.tech/canvas_319.png?t...

One mole of an ideal monoatomic gas at temperature expands slowly according to the law constant. If the final temperature is heat supplied to the gas is

A

2RT02 \mathrm { RT } _ { 0 }

B

RT0\mathrm { RT } _ { 0 }

C

D

12RT0\frac { 1 } { 2 } \mathrm { RT } _ { 0 }

Answer

2RT02 \mathrm { RT } _ { 0 }

Explanation

Solution

: In a process PVxP V ^ { x }= constant, molar heat capacity is given by C=Rγ−1+R1−xC = \frac { R } { \gamma - 1 } + \frac { R } { 1 - x }

As the process is PV=\frac { P } { V } =constant

i.e. PV−1=P V ^ { - 1 } =constant, therefore, x=−1x = - 1

for an ideal monatomic gas, γ=53\gamma = \frac { 5 } { 3 }

∴C=R53−1+R1−(−1)=32R+R2=2R\therefore C = \frac { R } { \frac { 5 } { 3 } - 1 } + \frac { R } { 1 - ( - 1 ) } = \frac { 3 } { 2 } R + \frac { R } { 2 } = 2 R

ΔQ=nC(ΔT)=1(2R)(2T0−T0)=2RT0\Delta Q = n C ( \Delta T ) = 1 ( 2 R ) \left( 2 T _ { 0 } - T _ { 0 } \right) = 2 R T _ { 0 }