152 mL of a gas at STP was taken to 20-C and 729 mm pressure- What was the change in volume of the gas-

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Standard XII

Physics

Temperature

152 mL of a...

Question

152 $m L$ of a gas at $S T P$ was taken to $20^{\circ} C$ and 729 $m m$ pressure. What was the change in volume of the gas?

V_{2} - V_{1} = 9 m L

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V_{2} - V_{1} = 18 m L

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V_{2} - V_{1} = 36 m L

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None of these

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Solution

The correct option is B $V_{2} - V_{1} = 18 m L$
Given
$V_{1} = 152 m L$
$P_{1} = 1 a t m = 760 m m$
$T_{1} = 273.15 K$
$V_{2} = ?$
$P_{2} = 729 m m$
$T_{2} = 20^{\circ} C$ or $293.15 K$
$∴ V_{2} = \frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}} = \frac{760 \times 152 \times 293.15}{273.15 \times 729} = 170 m L$
$∴ C h a n g e i n v o l u m e = V_{2} - V_{1} = 170 - 152 = 18 m L$

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152 mL of a gas at STP was taken to 20∘C and 729 mm pressure. What was the change in volume of the gas?

The correct option is B V2−V1=18mLGivenV1=152mLP1=1atm=760mmT1=273.15KV2=?P2=729mmT2=20∘C or 293.15K∴V2=P1V1T1×T2P2=760×152×293.15273.15×729=170mL∴Changeinvolume=V2−V1=170−152=18mL

152 $m L$ of a gas at $S T P$ was taken to $20^{\circ} C$ and 729 $m m$ pressure. What was the change in volume of the gas?