1 mole of gas expands isothermally at `37^(@)C`. The amount of heat is absorbed by it until its volume doubled is `(R= 8.31 J mol^(-1) K^(-1))`

Options

A. 411.25 cal
B. 418.50 cal
C. 420.25 cal
D. 425.40 cal

Answer

D

Explanation

Here `T=37^(@)c=37+273=310 K V_(2)=2V`
work done in siothermal process is
`W=nRTln(V_(2))/(V_(1))=1xx8.31xx310xxln(2V_(1))/(V_(1))`
`=8.31xx310xxln2=1.786xx10^(3)`
Amount of heat absorbedc =`(1.786xx10^(3))/(4.2)` cal =425.4 cal