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A gas is heated through $1^{\large\circ}C$ in a closed vessel and so the pressure increases by $0.4\%$ . The initial temperature of the gas was

$(a)\;250^{\large\circ}C\qquad(b)\;100^{\large\circ}C\qquad(c)\;-75^{\large\circ}C\qquad(d)\;-23{\large\circ}C$

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A)
Let $T_1 = T and\; P_1 = P$
$T_2$ = (T + 1)
$P_2 = P + \large\frac{0.4P}{100}$
$= \large\frac{100.4}{100}P$
Using
$\large\frac{P_1V_1}{T_1} = \large\frac{P_2V_2}{T_2}$
$\large\frac{PV}{T} = \large\frac{100.4P\times V}{100\times(T+1)}$
$\therefore T = 250K = -23^{\large\circ}C$
Hence answer is (d)
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