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

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Solution

let us consider the equation
$T = \frac{P V}{n R}$
Given that:
$P = 2.05 a t m$
$V = 2 L$
$n = \frac{1 g}{4 g m o l^{- 1}} = 0.25 m o l$
$R = 0.0821 L a t m K^{- 1} m o l^{- 1}$
Putting all values in above equation,
$T = \frac{2.05 \times 2}{0.25 \times 0.0821}$
Implies that
$= 199.76 K$ .

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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?

let us consider the equation T=PVnRGiven that:P=2.05atmV=2Ln=1g4gmol−1=0.25molR=0.0821LatmK−1mol−1Putting all values in above equation,T=2.05×20.25×0.0821Implies that=199.76K.

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

let us consider the equation
$T = \frac{P V}{n R}$
Given that:
$P = 2.05 a t m$
$V = 2 L$
$n = \frac{1 g}{4 g m o l^{- 1}} = 0.25 m o l$
$R = 0.0821 L a t m K^{- 1} m o l^{- 1}$
Putting all values in above equation,
$T = \frac{2.05 \times 2}{0.25 \times 0.0821}$
Implies that
$= 199.76 K$ .