(a) 3 cal

(b) 4 cal

(c) 6 cal

(d) 9 cal

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Answer: 6 cal

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., $C_V$ = 4 cal

$\therefore C_P = C_V + R$ = 4 cal + 2 cal = 6 cal

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...