At constant temperature, the pressure of $22.4 d m^{3}$ volume of an ideal gas was increased form 105 kPa to 210 kPa. the new volume will be?

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Solution

Given $P_{1} = 10; V_{1} = 22.4 d m^{3}; P_{2} = 210 K P a; V_{2} = ?$
$P \propto \frac{1}{V}$
$\Rightarrow P_{1} V_{1} = P_{2} V_{2}$
$\Rightarrow V_{2} = \frac{P_{1} V_{1}}{P_{2}}$
$= \frac{105 \times 22.4}{210}$
$= 11.2 d m^{3}$ .

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