Question

Calculate the density of br2(g) at 59.9oc and 1.00 atm pressure

Answer 1

Answer : Density = 11.694 g/L

Explanation : To find out the density of gas of
`Br_(2)` we can use the ideal gas equation

PV = nRT ;

where P is pressure in atm,
V is volume of gas,
R is gas constant,
n is no. of moles of gas and T is temperature in K

On modifying this equation to find density we get;

n=m/M where m is mass of gas and M is molecular mass of gas

but m/V is density so modifying the above equation we get;
m/V = (M X P) / (R X T)

on substituting the given values we get;
m/V = (319.616 X 1) / ( 0.0821 X 332.9)

m/V = 11.694 g/L

hence density = 11.694 g/L

Answer 2

Answer : the density of the Br₂ at 59.9 °C and 1 atm is 5.85 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 = 1 atm = 101325 pa
d = ?
R = 8.314 J mol⁻¹ K⁻¹
T = (59.9 273) K = 332.9 K
M = 159.808 g/mol = 159.8 x 10⁻³ kg/mol

By substitution,

101325 Pa = (d x 8.314 J mol⁻¹ K⁻¹ x 332.9 K) / 159.8 x 10⁻³ kg/mol

d = (101325 Pa x 159.8 x 10⁻³ kg/mol) / (8.314 J mol⁻¹ K⁻¹ x 332.9 K)
d = 5.85 kg m⁻³

Hence, the density of the Br₂ at 59.9 °C and 1 atm is 5.85 kg m⁻³.

Assumption made is "Br₂ gas has an ideal gas behavior".