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

                  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)

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

answer

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