A gas is collected at a pressure of 95 cm of $H g$ and a temperature of $50^{0} C$ . To what temperature should it be cooled so that occupies a volume which is 80% of its original volume when the pressure of the gas is 90 cm of $H g$ ,

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Solution

$76 c m = 1 a t m$
$∴ 95 c m = \frac{5}{4} a t m = 1.25 a t m$
Now $T = 50 c = 323 k$
cet the no. of mole $= 1$
$P V = n R T$
$1.25 \times V = 1 \times \frac{1}{12} \times 323$
$V = \frac{323}{12 \times 1.25}$
$= 21.533$
Now
New $V = 80 %$ of old $V$
$∴ n e w v = 17.2264$
New $P = \frac{90}{76}$
$= P V = n R T$
$= \frac{90}{76} \times 17.226 = 1 \times \frac{1}{12} \times T$
$= T (90 \times 17.226 \times 12) \div 76$
$= 244.79$
$= 244.8 K$

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A gas is collected at a pressure of 95 cm of Hg and a temperature of 500C. To what temperature should it be cooled so that occupies a volume which is 80% of its original volume when the pressure of the gas is 90 cm of Hg,

76cm=1atm∴95cm=54atm=1.25atmNow T=50c=323kcet the no. of mole =1PV=nRT1.25×V=1×112×323V=32312×1.25=21.533NowNew V=80% of old V∴newv=17.2264New P=9076=PV=nRT=9076×17.226=1×112×T=T(90×17.226×12)÷76=244.79=244.8K

A gas is collected at a pressure of 95 cm of $H g$ and a temperature of $50^{0} C$ . To what temperature should it be cooled so that occupies a volume which is 80% of its original volume when the pressure of the gas is 90 cm of $H g$ ,