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

Answer

D

Explanation

For one mole of an ideal gas,
`C_(P) - C_(V) = R`
or `C_(V) = C_(V) + R`
Thus, we must find the molar heat capacity at constant volume `(C_(V))`.
`n_(gas) = ("Volume" (L) "at" STP)/(22.4 L mol^(-1)) = (4.48 L)/(22.4 L mol^(-1)) = 0.2 mol`
According to thermodynamics,
`q = nC_(V) Delta T`
or `C_(V) = (q)/(n Delta T) = (12 cal)/((0.2 mol) (15^(@))`
` = 4 cal mol^(-1) .^(@)C^(-1)`
Substituting this value in the above equation, we get
`C_(P) = C_(V) + R`
`= (4 cal mol^(-1) .^(@)C^(-1)) + (2 cal mol^(-1) .^(@)C^(-1))`
`= 6 cal mol^(-1) .^(@)C^(-1))`