`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

No. of moles in 4.48 L of ideal gas at STP `= ( 4.48 )/( 22.4) = 0.2`
Thus, to raise the temperature of0.2 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) = 4 cal` i.e., `C_(v) = 4 cal`
`:. C_(p) = C_(v) + R = 4+2 cal = 6 cal`.