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The prssure in a monoatomic gas increase...

The prssure in a monoatomic gas increases lineraly from `4 xx 10^(5) Nm^(-2) to 8 xx 10^(5)Nm^(-2)` when its volume increases form `0.2 m^(3) to 0.5 m^(3)`. Calculate the following:
(a) work done by the gas. (2) increase in the internal energy

Text Solution

Verified by Experts

(a) As here pressure is varying lineraly with volume, work done by the gas
`W = int PdV =` area under `P-V` curve
`W = P_(1) (V_(F) -V_(1)) +(1)/(2) (P_(F) -P_(1)) xx (V_(F) -V_(1))`
i.e., `W = 4 xx 10^(5) xx 0.3 +(1)/(2) xx4 xx 10^(5) xx 0.3`
i.e., `W = 1.8 xx 10^(5) J`
(b) The change in internal energy of a gas is given by
`DeltaU = nC_(v)DeltaT = (nRDeltaT)/((gamma-1)) = ((P_(F)V_(F)-P_(1)V_(1)))/((gamma-1))`
As the gas is monoatomic `gamma = (5//3)`
So, `DeltaU = (10^(5) (8 xx 0.5 - 4 xx 0.2))/([(5//3)-1]) = (3)/(2) xx 10^(5) (4-0.8)`
i.e., `DeltaU = 4.8 xx 10^(5) J`
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