Calculate the pressure of 3.2 moles of an ideal gas that occupies a volume of 87 m3 at a temperature of 312 K.

The correct answer is 95.36 Pa.

Explanation:

It is given that the moles of an ideal gas, n = 3.2 moles

Volume of an ideal gas, V = 87 m³

Temperature, T = 312 K

Pressure, P = x

The ideal gas equation, PV = nRT, here R is the gas constant, and at standard temperature and pressure, one mole of ideal gas holds 22.4 L volume, At STP, T = 0 degree C = 273 K

V = 22.4 L

moles, n = 1 mole

Gas constant, R = PV / nT

R = 1 atm * 22.4 L / 1 mole * 273 K

1 atm = 1.013 * 10⁵ Pa and 1L = 10⁻³ m³

R = 8.31 Pa. m³. mol⁻¹. K⁻¹

Now, there is a need to calculate pressure, P:

PV = nRT

P = 3.2 mol * 8.31 Pa. m³. mol⁻¹. K⁻¹ * 312 K / 87 m³

P = 8296.7 / 87

P = 95.36 Pa

P = 0.0009417 atm

Or,

P = 9.417 * 10⁻⁴ atm

Or,

P = 0.0954157 kPa

Or,

P = 0.715677 mmHg (Torr)

Explanation:

Data Given:

Moles = n = 3.2 mol

Temperature = T = 312 K

Pressure = P = ?

Volume = V = 87 m³ = 87000 L

Formula Used:

Let's assume that the gas is acting as an Ideal gas, the according to Ideal Gas Equation,

P V = n R T

where; R = Universal Gas Constant = 0.082057 atm. L. mol⁻¹. K⁻¹

Solving Equation for P,

P = n R T / V

Putting Values,

P = (3.2 mol * 0.082057 atm. L. mol⁻¹. K⁻¹ * 312 K) : 87000 L

P = 0.0009417 atm

Or,

P = 9.417 * 10⁻⁴ atm

Or,

P = 0.0954157 kPa

Or,

P = 0.715677 mmHg (Torr)