Six moles of `O_(2)` gas is heated from `20^(@)C` to `35^(@)C` at constant volume . If specific heat capacity at constant pressure is `8 " cal mol"^(-1)-K^(-1) and R=8.31"J mol"^(-1)-K^(-1)` , what is change in internal energy of gas ?

Options

A. 180 Cal
B. 300 cal
C. 360 cal
D. 540 cal

Answer

D

Explanation

Consider n moles of a gas which undergo isochoric process, i.e .V = Constant . Form first law of thermodynamics, `DeltaQ=DeltaW+DeltaU` …(i)
Here , `DeltaW=0` as V = constant
`DeltaQ=nCVDeltaT`
Subsitituting in Eq .(i) ,we get
`DeltaU=nC_(v)DeltaT ` ...(ii) Mayer's relation can be written as
`C_(p)-C_(v)=RrArrC_(v)=C_(p)_R` ...(iii)
From Eqs (ii)and(iii) , we have `DeltaU=n(C_(p)-R)DeltaT`
Given , `n=6,C_(p)=8 "cal mol"^(-1)-K^(-1)~~2"cal mol"^(-1)-K^(-1)`
Hence , `Delta U =6(8-2)(35-20)=6xx6xx15=540"cal"`