`4.48 L` of an ideal gas at `STP` requires 12 cal to raise its temperature by `15^(@)C` at constant volume. The `C_(P)` of the gas is

Options

A. `3cal`
B. `4cal`
C. `7cal`
D. `6 cal`

Answer

D

Explanation

No. of moles in `4.484` of ideal gas at STP
`=(4.48)/(22.4)=0.2`
Thus to riase the temperature of `0.2` mol of the ideal gas, through `15^(@)C` heat absorbed `=12 cal`.
`:. ` To raise the temperature of `1` mol of the gas through `1^(@)C`, heat absorbed `=(12)/(15)xx(1)/(0.2)=4cal`
i.e., `C_(V)=4cal`
`:. C_(P)=C_(V)+R=4+2cal=6cal`