Question

What is the pressure exerted by 5.00 moles of nitrogen gas contained in a 30.0 liter container at 25.0 °c?

Answer 1

Given: number of moles of N2 = 5
volume of container = 30 l
temperature = 25 oC = 298 K

Also, we know that R = gas consant = 0.082 L atm mol-1 K-1

Now from ideal gas equation we have,

PV = nRT
∴ P = nRT/V
= (5 X 0.082 X 298)/30
= 4.07 atm.

Thus, pressure exerted by gas is 4.07 atm

Answer 2

Let's assume that the given gas has an ideal gas behavior. So we can use ideal gas law,
PV = nRT
Where P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant (8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin (K).

P = ?
V = 30.0 L = 30.0 x 10⁻³ m³
n = 5.00 mol
R = 8.314 J mol⁻¹ K⁻¹
T = 25.0 + 273 K = 298 K

By applying the equation,
P x 30.0 x 10⁻³ m³ = 5.00 mol x 8.314 J mol⁻¹ K⁻¹ x 298 K
P = 412928.67 Pa
P = 4.13 x 10⁵ Pa

Hence, the pressure of the gas is 4.13 x 10⁵ Pa