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Question

# An ideal monatomic gas is at P0,V0. It is taken to final volume 2V0 when pressure is P0/2 in a process which is straight line on P-V diagram. Choose the correct statements.

A
Final temperature is equal to initial temperature.
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B
There is no change in internal energy.
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C
Work done by the gas is +P0V04
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D
Heat is absorbed in the process.
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Solution

## The correct options are A Final temperature is equal to initial temperature. B There is no change in internal energy. D Heat is absorbed in the process.(a) We have PV=nRTIf T1 and T2 are initial and final temperatures,PoVo=nRT1 and Po22Vo=nRT2⟹T1=T2(b) T1=T2⟹ΔU=0(c) Work done = area under the curveW=12×Po/2×(2Vo−Vo)+Po/2×(2Vo−Vo) =14×Po×Vo+12×Po×Vo =34×Po×Vo(d) We have, ΔQ−ΔW=ΔUSince ΔU=0 and ΔW is positive (i.e.s work is done by the system),ΔQ>0, heat is absorbed in the process.

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