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To raise the temperature of 4.48 litres of an ideal gas at STP by $15^\circ C$ it requires 12.0 calories. The $C_P$ of the gas is

(a) 3 cal
(b) 4 cal
(c) 6 cal
(d) 9 cal

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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
answered Feb 27, 2014 by mosymeow_1

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