1/2 mole oxygen gas occupies 44.8 L at temperature 273.15 K. The pressure of the oxygen gas is .

0.25 atm

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0.5 atm

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4 atm

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Solution

The correct option is A 0.25 atm
Given V= 44.8 L, T = 273.15 K, n = 1/2 and R = 0.082 $L i t r e a t m K^{- 1} m o l^{- 1}$

The ideal gas equation is PV = nRT
where P = Pressure,
V = Volume,
n = Number of moles of the gas,
R = Universal gas constant,
T = Temperature

By substituting the given values in the equation we get
P x 44.8 = 1/2 x 0.082 x 273.15
P = 0.25 atm

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