One litre of a gas weighs $2 g$ at $300 K$ and $1 a t m$ pressure. If the pressure is made $0.75 a t m$ at which of the following temperatures will one litre of the same gas, weigh one gram?

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Solution

$P V = n R T$ (Ideal gas equation.)
$⟹ \frac{P_{1} V_{1}}{P_{2} V_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}} ⟹ \frac{1 \times 1}{0.75 \times 1} = \frac{\frac{2}{M_{w}} \times 300}{\frac{1}{M_{w}} \times T}$
$⟹ T = \frac{2 \times 300 \times 0.75}{1} = 450 K$
$∴$ $Temperature = 450 K .$

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Q. One litre of a gas weighs 2g at

300 K

and

1

atm. If the pressure is made

0.75

atm at what temperature will one litre of the same gas weigh one gram?

Q. 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 ?

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2

g at

300

K and

1

atm pressure. If the pressure is made

0.75

atm, at which temperature will one litre of the same gas weighs

1

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One litre of a gas weighs 2g at 300K and 1atm pressure. If the pressure is made 0.75atm at which of the following temperatures will one litre of the same gas, weigh one gram?

PV=nRT (Ideal gas equation.)⟹P1V1P2V2=n1T1n2T2⟹1×10.75×1=2Mw×3001Mw×T⟹T=2×300×0.751=450K∴ Temperature =450K.

One litre of a gas weighs $2 g$ at $300 K$ and $1 a t m$ pressure. If the pressure is made $0.75 a t m$ at which of the following temperatures will one litre of the same gas, weigh one gram?

$P V = n R T$ (Ideal gas equation.)
$⟹ \frac{P_{1} V_{1}}{P_{2} V_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}} ⟹ \frac{1 \times 1}{0.75 \times 1} = \frac{\frac{2}{M_{w}} \times 300}{\frac{1}{M_{w}} \times T}$
$⟹ T = \frac{2 \times 300 \times 0.75}{1} = 450 K$
$∴$ $Temperature = 450 K .$