One mole of air $(C_{v} = \frac{5 R}{2})$ is confined at atmospheric pressure in a cylinder with a piston at 0 $° C$ . The initial volume occupied of gas is V. After the equivalent of 13200 J of heat is transferred to it, the volume of gas $V_{f}$ is nearly Litre. (1 atm = $10^{5} N / m^{2}$ )

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Solution

$P = 1 a t m = 10^{5} N / m^{2}$ $T = 0^{0} C = 273 K$ $V = \frac{n R T}{P} = \frac{1 \times 8.3 \times 273}{10^{5}} = 0.0227 m^{3} = 22.7 L$
$C_{V} = \frac{5}{2} R, C_{P} = \frac{7}{2} R$
Heat transferred $Δ Q = n C_{P} Δ T = \frac{n (7 R)}{2} Δ T = 13200 J$
Work done $n R Δ T = (13200 \times \frac{2}{7})$
$= P (V_{f} - V_{i}) = 3771$
$V_{f} - V_{i} = 3771 \times 10^{- 5} = 0.0377 m^{3}$ $V_{f} = V_{i} + 37.7 L =$ $22.7 L + 37.7 L$
= $60.4 L$

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