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The pressure in monoatomic gas increases...

The pressure in monoatomic gas increases linearly from ` =4xx10^(5) Nm^(-2)` to ` 8xx10^(+5) Nm^(-2)` when its volume increases from `0.2m^(3)` to `0.5m^(3)`. Calculate .
(i) Work done by the gas ,
(ii) Increase in the internal energy,
(iii) Amount of heat supplied,
(iv) Molar heat capacity of the gas `R= 8.31J mol^(-1) K^(-1)`

Text Solution

Verified by Experts

`P_(1)= 4xx10^(5) Nm^(-2) P_(2)= 8xx10^(+5) Nm^(-2), V_(1) = 0.2m^(3), V_(2) = 0.5m^(3)`
(i) Work done by the gas = Area under `P-V` graph (Area`ABCDEA`)
=`1/2 (AE+AD) xxAC = (1)/(2) (4xx10^(5)+ 8xx10^(5))xx(0.5-0.2)`
=` 1/2 xx12xx10^(5)xx0.3 = 1.8xx10^(5)J`
(ii) Increase in internal energy `DeltaU=C_(v)(T_(2)-T_(1)) = (C_(V))/(R)R(T_(2)- T_(1)) = (C_(v))/(R)(P_(2)V_(2)-P_(1)V_(1))`
For monoatomic gas
`C_(v) = (3)/(2)R therefore DeltaU=(3)/(2)[(8xx10^(5)xx0.5)-(4xx10^(5) xx 0.2)] =(3)/(2)[4xx10^(5)-0.8xx10^(5)] = 4.8xx10^(5)J`
(iii) ` Q=DeltaU + W=4.8 xx 10^(5) + 1.8 xx 10^(5) = 6.6 xx 10^(5)J`
(iv) `C= (Q)/(etaDeltaT) = (QR)/(etaRDeltaT) = (QR)/(eta(P_(2)V_(2)- P_(1)V_(1))) = (6.6xx10^(5)xx8.31)/(1xx3.2xx10^(5)) = 17.14J//"mole" K`
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