The ratio of specific heats of a gas is 1.41. If at `NTP_(1)` the volume occupied by the gas is `0.0224m^(3),` then calculate the molar specific heats at constant volume and pressure.

Explanation

Given `sqrt(2) = 1.41`
`V_(0)=0.0224m^(3),P_(o)=1.013xx10^(5)Nm^(-2)`
`T_(o)=273`
We know that `R=(P_(o)V_(o))/(T_(o))=(1.013xx10^(5)xx0.0224)/(273)`
`i.e. " " R=8.312J "moles"^(-1)K^(-1)`
Also `lambda=(C_(p))/(C_(v))i.e.C_(p)=1.41C_(Y)`
From Mayer's equation `C_(p)-C_(Y)=R`
`i.e.(1.41-1)C_(Y)=R`
`therefore " " `C_(Y)=(R)/(0.41)`=(8.312)/(0.41)=20.27J"mole"^(-1)K_(1)`
Hence `C_(p)=1.41xx20.27=28.58j"mole"^(-1)K^(-1)`