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At 27^@C two moles of an ideal monatomic...

At `27^@C` two moles of an ideal monatomic gas occupy a volume V. The gas expands adiabatically to a volume `2V`. Calculate
(a) final temperature of the gas
(b) change in its internal energy and
(c) the work done by the gas during the process. [ `R=8.31J//mol-K`]

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
(a) In case of adiabatic change
`TV^(gamma-1)=` constant
So that
`T_1V_1^(gamma-1)=T_2V_2^(gamma-1)` with `gamma=(5/3)`
i.e. `300xxV^(2//3)=T(2V)^(2//3)`
or `T=(300)/((2)^(2//3))=189K`
(b) As `DeltaU=nC_VDeltaT=n(3/2R)DeltaT
So, `DeltaU=2xx(3/2)xx8.31(189-300)`
`=-2767.23J`
Negative sign means internal energy will decrease.
(c) According to first law of thermodynamics
`Q=DeltaU+DeltaW`
And as for adiabatic change `DeltaQ=0`
`DeltaW=-DeltaU=2767.23J`
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