1litre of a gas weights 2g at 300K and 1atm pressure.If the pressure is made 0.75atm at which temperature will 1L of same gas weighs 1g ?

600k

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900K

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800K

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450K

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Solution

The correct option is D 450K
$P_{1}$ = 1 atm ; $W_{1}$ = 2g

$P_{2}$ = 0.75 atm ; $W_{2}$ = 1 g

$T_{1}$ = 300 K ; $V_{1}$ = 1 L

$T_{2}$ = ? ; $V_{2}$ = 1 L

So we know that,

$P_{1} V_{1} / P_{2} V_{2} = W_{1} T_{1} / W_{2} T_{2}$

Substituting in the above formula we get,

$= 1 \times 1 / 0.75 \times 1 = 2 \times 300 / 1 \times T_{2}$

$= 1 / 0.75 = 600 / T_{2}$

Cross Multiplying we get,

$= T_{2} \times 1 = 600 \times 0.75$

$= T_{2} = 450$ K

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1litre of a gas weights 2g at 300K and 1atm pressure.If the pressure is made 0.75atm at which temperature will 1L of same gas weighs 1g ?

The correct option is D 450KP1 = 1 atm ; W1 = 2gP2 = 0.75 atm ; W2 = 1 gT1 = 300 K ; V1 = 1 LT2 = ? ; V2 = 1 LSo we know that,P1V1/P2V2=W1T1/W2T2Substituting in the above formula we get,=1×1/0.75×1=2×300/1×T2=1/0.75=600/T2Cross Multiplying we get,=T2×1=600×0.75=T2=450 K