A diatomic ideal gas is heated at constant volume until its pressure becomes three times. It is again heated at constant pressure until its volume is doubled. Find the molar heat capacity for the whole process.

Answer

A, C

Explanation

First process
`V=const ant`
`:.` `ppropT`
Pressure becomes three times.
Therefore, temperature also becomes three times.
`(T_f=3T_i)`
Second process
`p=const ant`
`:.` `VpropT`
Volume is doubled. So, temperature also becomes two times. `(T_f=6T_i)`
Now, `C=(Q)/(nDeltaT)=(Q_1+Q_2)/(nDeltaT)`
`=(nC_VDeltaT_1+nC_pDeltaT_2)/(nDeltaT)`
`=((5/2R)(3T_i-T_i)+(7/2R)(6T_i-3T_i))/(6T_i-T_i)`
`=3.1R`