Question

What is the pressure in millimeters of mercury of 0.0105 mol m o l of helium gas with a volume of 223 mL m L at 50. ∘C ∘ C ? (Hint: You must convert each quantity into the correct units (L L , atm a t m , mol m o l , and K K ) before substituting into the ideal gas law.)

Answer 1

The pressure of helium gas in the given condition is 948.45 mm of Hg.

Explanation:

As per ideal gas law, the product of pressure and volume will be equal to the product of number of moles, gas constant and temperature of gas molecules. This is formed by the combination of three basic laws of kinetic theory of gases.

$PV = nRT$

As the pressure P is unknown, but the volume V, temperature T and number of moles n is given for helium gas.

R = 8.314 J mol⁻¹ K⁻¹ = 62.363 mmHg L mol⁻¹ K⁻¹.

Then, pressure can be found as

$P = \frac{nRT}{V}$

As T = 50°C = 50 + 273 K = 323 K and volume V = 223 mL = 0.223 L and n = 0.0105 mol

Then, $P = \frac{0.0105*62.363*323}{0.223}=948.45 mm of Hg$

So, the pressure of helium gas in the given condition is 948.45 mm of Hg.