Question

4.48 litres of an ideal gas at STP requires 12.0 calories to raise its temperature by 15$\circ$C at constant volume. The $C_p$ of the gas is

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

Answer 1

Answer: (D)

6 cal

Answer 2

No. of moles in 4-484 of ideal gas at STP = $\frac{4.48}{22.4}$ = 0.2 Thus to raise the temperature of 0-2 mol of the ideal gas, through 15$\circ$C heat absorbed = 12 cal. $\therefore$ To raise the temperature of 1 mol of the gas through 1$\circ$C, heat absorbed = \frac{12}{15} \times \frac{1}{0.2 = 4 cal i.e., Cv = 4 cal CP = Cv + R = 4 + 2 cal = 6 cal