$(a)\;1m^3\qquad(b)\;2m^3\qquad(c)\;3m^3\qquad(d)\;4m^3$

Given

$P_2 = \large\frac{1}{2}P_1$

$T_2 = 2T_1$

$V_1 = 1.0m^3$

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)

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