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)`.

Explanation

The gas is monatiomic, `C_V=(3R)/(2)`
Let number of moles are `n`.
`P_1=4xx10^5(N)/(m^2)`,`V_1=0.2m^3`
`P_2=8xx10^5(N)/(m^2)`,`V_2=0.5m^3`
`PV=nRTimpliesT=(PV)/(nR)`
`T_1=(P_1V_1)/(nR)=(4xx10^5xx0.2)/(nR)=(8xx10^4)/(nR)`
`T_2=(P_2V_2)/(nR)=(8xx10^5xx0.5)/(nR)=(40xx10^4)/(nR)`
(a) To calculate work done, sketch `P-V` diagram
`DeltaW_(1rarr2)=`area below `1rarr2`
`=(1)/(2)(4xx10^5+8xx10^5)(0.5-0.2)`
`=6xx10^5xx0.3=18xx10^4J`
(b) `DeltaU=nC_VDeltaT=nC_V(T-2-T_1)`
`=n.(3R)/(2)((40xx10^4)/(nR)-(8xx10^4)/(nR))=(3)/(2)xx32xx10^4`
`=48xx10^5J`
(c) `DeltaQ=DeltaU+DeltaW=18xx10^4+48xx10^4=66xx10^4J`
(d) `DeltaQ=nCDeltaT`
`C=(DeltaQ)/(nDeltaT)=(66xx10^4)/(nxx(32xx10^4)/(nR))=(66R)/(32)=(33)/(16)xx8.3`
`=17.1J//"mole K"`