One litre of a gas weighs 2 g 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?

450 K

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800 K

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600 K

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900 K

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Solution

The correct option is A 450 K
Given $P_{1} = 1 a t m P_{2} = 0.75 a t m, V_{1} = 1 L V_{2} = 1 L$
$m_{1} = 2 g m_{2} = 1 g$ $T_{1} = 300 K T_{2} = ?$
$R = 0.082 L$

According to Ideal gas law,

$P V = n R T$

Where $n = \frac{m}{M}$ (where $m =$ mass & $M =$ molar mass)

So, $P_{1} V_{1} = \frac{m_{1}}{M} R T_{1}$

$M = \frac{m_{1} R T_{1}}{P_{1} V_{1}} = \frac{2 \times 0.082 \times 300}{1 \times 1} = 49.2 g / m o l$

$∴ T_{2} = \frac{P_{2} V_{2} M}{m_{2} R} = \frac{0.75 \times 1 \times 49.2}{1 \times 0.082} = \frac{36.9}{0.082}$

$= 450 K$

Hence, option A is correct.

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