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The pressure of one gram mole of a monoa...

The pressure of one gram mole of a monoatomic gas increases linearly from `4xx10^(5)N//M^(2)` to `8xx10^(5)N//m^(2)`. Calculate
(i) Work done by the gas,
(ii) increase in internal energy, (iii) amount of heat supplied,
(iv) molar heat capacity of the gas.
Take `R=8.31J mol e^(-1)K^(-1)`.

Text Solution

Verified by Experts


`P_(1)= 4xx10^(5)N//m^(2)`
`V_(1)=0.2m^(3)`
`P_(2)=8xx10^(5)N//m^(2)`
`V_(2)= 0.5m^(3)`
(i) Work done by the gas
W = area under P-V curve
`= "area of rect". ACDE+"area of" DeltaABC`
`=P_(1)(V_(2)-V_(1))+1/2(P_(2)-P_(1))(V_(2)-V_(1))`
`=(V_(2)-V_(1))[P_(1)+1/2P_(2)-1/2P_(1)]`
`=1/2(V_(2)-V_(1))(P_(1)+P_(2))`
`=1/2(0.5-0.2)(4xx10^(5)+8xx10^(5))`
`W=1/2xx0.3xx12xx10^(5)= 1.8xx10^(5)= 1.8xx10^(5)J`
(ii) Change in internal energy of gas
`dU= n C_(v).dT`
`=n((R)/(gamma-1))xxdT=(n(P_(2)V_(2)-P_(1)V_(1)))/(gamma-1)`
`=(1(8xx10^(5)xx0.5-4xx10^(5)xx0.2))/(5//3-1)`
`dU=3.2xx10^(5)xx3/2= 4.8xx10^(5)J`
(iii) From first law of thermodynamics
`dQ=dU+dW`
`= 4.8xx10^(5)+1.8xx10^(5)= 6.6xx10^(5)J`
(iv) `C=(dQ)/(ndT)=(dQxxR)/(n(P_(2)V_(2)-P_(1)V_(1)))`
`=(6.6xx10^(5)xx8.31)/(1(8xx10^(5)xx0.5-4xx10^(5)xx0.2)`
`C=17.14J mol e^(-1)K^(-1)`
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