States and Properties of Matter - The Gas Laws (2)

Your feedback on these self-help problems is appreciated. Click here to send an e-mail.

For on-line help with this topic, see the chemcal module "Behaviour of Gases".

Shortcut to Questions

Q: 1 2 3 4 5

1

The density of a gas at 298 K and 101 kPa is 1.70 gram litre-1. Calculate its molar mass.
2

An unknown compound is analysed and its empirical formula determined to be CHCl. The vapour density measured at 380 K and 101 kPa is 3.10 gram litre-1. What is the molecular formula?

3

A fluorocarbon has the molecular formula C4F8. Calculate the density of this gas at 400 K and 202 kPa?

4

At what temperature will the density of methane gas be 0.65 gram litre-1 if the pressure is 101 kPa?
5

The density of a certain gas is 1.375 gram litre-1 at 273 K and 101.3 kPa. Calculate its density at 303 K and 98.0 kPa.

The Gas Laws 2 (Answers)

  

1

The problems in this section combine the Ideal Gas Equation (IGE) with the definition of a mole = mass / molar mass.

PV = nRT = (mass / molar mass)RT

The density is 1.70 g litre-1

Take a volume of 1.00 L

Therefore, mass = 1.70 g.


P = 101 kPa, T = 298 K and moles, n, = 1.70 / molar mass

Substitute in the IGE

101 x 1.00 = (1.70 / molar mass) x 8.314 x 298

[Note: for P given in kPa and V in L, R has the value 8.314 J K-1 mol-1.T is always in Kelvin.]

molar mass = (1.70 x 8.314 x 298) / (101 x 1.00) = 41.7 g mol-1

 

2

The empirical formula is CHCl,

so its empirical formula weight = 48.5

PV = nRT = (mass / molar mass)RT

The density is 3.10 g litre-1
Take a volume of 1.00 L

Therefore, mass = 3.10 g

Substitute in the IGE:
P = 101 kPa, T = 380 K and moles, n, = (3.10 / molar mass)

101 x 1.00 = (3.10 / molar mass) x 8.314 x 380

molar mass = (3.10 x 8.314 x 380) / (101 x 1.00) = 97.0 g mol-1

Therefore molar weight = 97.0 = 2 x empirical formula weight
and molecular formula is C2H2Cl2.
3

The molar mass of C4F8 = 200.0 g mol-1.

PV = nRT = (mass / molar mass)RT

Take 1.00 L of the gas.
Let m = the mass of the gas present in 1.00 L at 400 K and 202 kPa.

Therefore moles of gas at these conditions = m / molar mass

= m / 200.0 mol.

The density at these conditions = mass / V = (m / 1.00) g L-1
To calculate m and thence density, substitute in the Ideal Gas Equation with

P = 202 kPa and T = 400 K:

202 x 1.00 = (m / 200.0) x 8.314 x 400


m = (202 x 1.00 x 200) / (8.314 x 400) = 12.1 g

As V = 1.00 L, therefore density = 12.1 g L-1.

4

The molar mass of methane, CH4, = 16.05 g mol-1

Density = 0.65 g L-1, so for a volume = 1.00 L, mass = 0.65 g,
and number of moles in 1.00 L = 0.65 / 16.05 = 0.04050 mol

PV = nRT
101 x 1.00 = 0.04050 x 8.314 x T

Therefore, T = (101 x 1.00) / (0.04050 x 8.314) = 300 K
[Note: T is always in Kelvin in the Ideal Gas Equation]

5

From the first set of data, the molar mass of the gas can be calculated:
P = 101.3 kPa, T = 273 K, d = 1.375 g L-1

Therefore, for V = 1.00 L, mass of gas = 1.375 g

Let MW be the molar mass of the gas

Then number of moles present = n = (1.375 / MW)

Substitute in the ideal gas equation:

101.3 x 1.00 = (1.375 / MW) x 8.313 x 273

Therefore MW = (1.375 x 8.314 x 273) / (101.3 x 1.00) = 30.8

Using this result, the mass (m) of 1.00 L of the gas at the new conditions can be deduced by again using the IGE:

P = 98.0 kPa, T = 303 K, moles of gas, n, = (m / 30.8)


98.0 x 1.00 = (m / 30.8) x 8.314 x 303


m = (98.0 x 1.00 x 30.8) / (8.314 x 303) = 1.199 g

As this is the mass of gas in 1.00 L, then density = m / V = 1.199 g L-1.