What volume is occupied by 0.109 molmol of helium gas at a pressure of 0.98 atmatm and a temperature of 307 K

2.8 L

Explanation:

From the question given above, the following data were obtained:

Number of mole (n) = 0.109 mole

Pressure (P) = 0.98 atm

Temperature (T) = 307 K

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

The volume of the helium gas can be obtained by using the ideal gas equation as follow:

PV = nRT

0.98 × V = 0.109 × 0.0821 × 307

0.98 × V = 2.7473123

Divide both side by 0.98

V = 2.7473123 / 0.98

V = 2.8 L

Thus, the volume of the helium gas is 2.8 L.

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