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A certain amount of an ideal gas occupies a volume of $1.0m^3$ at a given temperature and pressure. What would be its volume after reducing its pressure to half the initial value and raising the temperature twice the initial value?


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$P_2 = \large\frac{1}{2}P_1$
$T_2 = 2T_1$
$V_1 = 1.0m^3$
Now $\large\frac{P_1\times1.0}{T_1} = \large\frac{1\times P_1\times V_2}{2\times2\times T_1}$
$\Rightarrow P_1\times1.0 = \large\frac{1\times P_1\times V_2}{4}$
$\Rightarrow 1.0 = \large\frac{1\times V_2}{4}$
$ 4 = V_2$
$V_2 = 4m^3$
Hence answer is (d)
answered Feb 15, 2014 by sharmaaparna1

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