Question

To raise the temperature of 4.48 litres of an ideal gas at STP by 15∘C it requires 12.0 calories. The CP of the gas is

Answer 1

No. of moles in 4.484 of ideal gas at STP = 4.4822.4=0.24.4822.4=0.2
Thus to raise the temperature of 0.2 mol of the ideal gas, through 15∘C15∘C heat absorbed = 12 cal.
∴∴ to raise the temperature of 1 mol of the gas through 1∘C1∘C, heat absorbed = 1215×10.2=41215×10.2=4 cal
i.e., CVCV = 4 cal
∴CP=CV+R∴CP=CV+R = 4 cal + 2 cal = 6 cal