What is the density of laughing gas, dinitrogen monoxide, n2o, at a temperature of 325 k and a pressure of 113.0 kpa?

Answer : the density of the N₂O at 325 K and 113.0 kPa  is 1.84 kg m⁻³.

Explanation : 

Density (kg/m³) = mass (kg) / Volume (m³)
 d = m/V                 (1)

Ideal gas law,
PV = nRT               (2)

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.

n = m/M          (3)

Where, n is number of moles, m is mass and M is molar mass.

From (2) and (3),
PV = (m/M) RT

By rearranging,
P = (m/VM)RT              (4)

From (1) and (4)
P = (dRT) / M

The given data,
P = 113.0 kPa = 113.0 x 10³ Pa
d = ?
R = 8.314 J mol⁻¹ K⁻¹
T = 325 K 
M = 44.0 g/mol  = 44.0 x 10⁻³ kg/mol

By substitution,

113.0 x 10³ Pa = (d x 8.314 J mol⁻¹ K⁻¹ x 325 K) / 44.0 x 10⁻³ kg/mol

             d = (113.0 x 10³ Pa x 44.0 x 10⁻³ kg/mol) / (8.314 J mol⁻¹ K⁻¹ x 325 K)
             d = 1.84 kg m⁻³

Hence, the density of the N₂O at 325 K and 113.0 kPa is 1.84 kg m⁻³.

Assumption made is "N₂O gas has an ideal gas behavior".

1 a

2 d

3 a

4 c

explanation: