A quantity of an ideal gas is collected in a graduated tube over the mercury in a barometer type arrangement. The volume of gas at $20^{o}$ C is $50$ ml and the level of mercury is $100$ mm above the outside of the mercury level. The atmospheric pressure is $750$ mm. Volume of gas at STP is :(Take $R = 0.083$ it. atm/K/mole).

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Solution

$P_{1} = 650$ torr, $V_{1} = 50 m l$ , $T_{1} = 20^{o} C = 293 K$
$P_{2} = 760$ torr, $V_{2} = ?$ , $T_{2} = 273 K$
Ideal gas Law: $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ rearranges to $V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$
$\Rightarrow V_{2} = \frac{(650) (50) (273)}{(760) (293)}$
$\Rightarrow 39.84 m l$
$\Rightarrow$ Volume of gas at STP is $39.84 m l$

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P1=650 torr, V1=50ml, T1=20oC=293KP2=760torr, V2=?, T2=273KIdeal gas Law: P1V1T1=P2V2T2 rearranges to V2=P1V1T2T1P2⇒V2=(650)(50)(273)(760)(293)⇒39.84ml⇒ Volume of gas at STP is 39.84ml