A certain mass of gas occupied a volume of 640 ml at a certain temperature and pressure. If the temperature is decreased by 40%, what will be the volume occupied by the same mass of gas under the same pressure?

256 m

l

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1600 m

l

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1066 m

l

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384 m

l

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Solution

The correct option is D 384 m $l$
We $P V = n R T$
for 1 mole and constant presssure
$V / T = R$ and R is a constant
So, $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
60% of $T_{1} = T_{2}$ $60 T_{1} / 100 = 0.6 T_{1}$ So,
$\frac{640}{T_{1}} = \frac{V_{2}}{0.6 T_{1}}$
$V_{2} = 384 m l$

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A certain mass of gas occupied a volume of 640 ml at a certain temperature and pressure. If the temperature is decreased by 40%, what will be the volume occupied by the same mass of gas under the same pressure?

The correct option is D 384 mlWe PV=nRTfor 1 mole and constant presssure V/T=R and R is a constant So, V1T1=V2T260% of T1=T2 60T1/100=0.6T1 So, 640T1=V20.6T1V2=384ml

The correct option is D 384 m $l$
We $P V = n R T$
for 1 mole and constant presssure
$V / T = R$ and R is a constant
So, $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
60% of $T_{1} = T_{2}$ $60 T_{1} / 100 = 0.6 T_{1}$ So,
$\frac{640}{T_{1}} = \frac{V_{2}}{0.6 T_{1}}$
$V_{2} = 384 m l$