$1 g$ of helium gas is confined in a two-litre flask under a pressure of $2.05 a t m$ . What is its temperature?

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Solution

Mass of $H e$ gas $= 1 g$
Volume $= 2 l i t$
$P = 2.05 a t m$
$n_{H e} = \frac{w}{M} = \frac{1}{4}$

Now,
$P V = n R T$

$T = \frac{P V}{R \times n} = \frac{2.05 \times 2}{0.0821 \times 1 / 4} = \frac{2.05 \times 8}{0.0821} = 200 K$

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1 g of helium gas is confined in a two-litre flask under a pressure of 2.05 atm. What is its temperature?

Mass of He gas =1gVolume =2 litP=2.05 atmnHe=wM=14Now,PV=nRTT=PVR×n=2.05×20.0821×1/4=2.05×80.0821=200 K

$1 g$ of helium gas is confined in a two-litre flask under a pressure of $2.05 a t m$ . What is its temperature?

Mass of $H e$ gas $= 1 g$
Volume $= 2 l i t$
$P = 2.05 a t m$
$n_{H e} = \frac{w}{M} = \frac{1}{4}$

Now,
$P V = n R T$

$T = \frac{P V}{R \times n} = \frac{2.05 \times 2}{0.0821 \times 1 / 4} = \frac{2.05 \times 8}{0.0821} = 200 K$