Numerical Problems based on combined Gas law or Gas equation & ideal gas equation

In this post, we will solve a few Numerical Problems based on the Combined Gas law (or Gas equation) & the Ideal gas equation.

The Formulas used to solve the numerical problems are:

  • PV / T = constant (k)
  • pV = nRT

Example 1

A sample of nitrogen occupies a volume of 1.0 L at a pressure of 0.5 bar at 40°C. Calculate the pressure if the gas is compressed to 0.225 mL at –6°C.

Solution: p1 = 0.5 bar

p2 = ?

V1 = 1.0 L

V2 = 0.225 mL = 0.225 × 10–3 L

T1 = 40+273 = 313 K

T2 = – 6 + 273 = 267 K

According to combined gas law equation,

p1V1/T1 = p2V2/T2

p2 = (p1V1T2)/(T1V2) = (.5 x 1x 267)/(313x 0.225 × 10–3) = 1895.6 bar

Example 2

At 25°C and 760 mm of Hg pressure, a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 10°C and volume of the gas is 640 mL.

Solution: p1=760 mm Hg

p2 = ?

V1=600 mL

V2 = 640 mL

T1=273 + 25 = 298 K

T2= 273 + 10 = 283 K

According to combined gas law equation,

p1V1/T1 = p2V2/T2

p2 = (p1V1T2)/(T1V2) = [(760 mm Hg) (600 mL) (283 K)] / [(298K)(640 mL) ] = 676.6 mm Hg

Example 3

Calculate the moles of hydrogen (H2) present in a 500 mL sample of hydrogen gas at a pressure of 1 bar and 27°C.

Solution:

1 atm = 1.01325 bar = 1.01325 × 105Pa = 101.325 kPa

or 1 bar = 0.987 atm

According to ideal gas equation,

pV = nRT

p = 1 bar = 1 atm = 101 x 10^3 Pa

V = 500 mL = 500 cm3 = 500 × 106 m3

T = 27 + 273 = 300 K,

R = 8.314 J mol–1K–1

Now, n = pV/(RT) = 101 x 10^3 x 500 × 106 / ( 8.314x 300) = 0.02 mole

Example 4

Calculate the volume occupied by 4.045 × 1023 molecules of oxygen at 27°C and having a pressure of 0.933 bar.

Solution: Here, number of molecules = 4.045 × 1023

p = 0.933 bar = 0.933 atm = 0.933 x 101 x 10^3 Pa

T = 27 + 273 = 300 K

R =8.314 J mol–1K–1

We will use pV = nRT.

V = nRT/p

Let us first calculate the number of moles, n.We know that number of moles n = No. of molecules / 6.022 x 1023

= 4.045 × 1023/ 6.022 x 1023 = 0.672 mol

V = nRT/p = 0.672 x 8.314 x 300 / ( 0.933 x 101 x 10^3) = 0.0177 m^3

Example 5

A discharge tube of 2 L capacity containing hydrogen gas was evacuated till the pressure inside is 1 × 10–5atm. If the tube is maintained at a temperature of 27°C, calculate the number of hydrogen molecules still present in the tube.

Solution:

Step I. To calculate the number of moles of hydrogen.

Here p = 1 × 10–5 atm = 1 × 10–5 x 101 x 103 Pa = 101 x 10-2 Pa

V = 2 litres =2 000 cm^3 = 2000 x 10-6 m3 = 2 x 10-3 m3

R = 8.314 J mol–1K–1

T = 27 + 273 = 300 K

n = ?

According to the gas equation,

pV = nRT

or n = pV/RT = 101 x 10-2 x 2 x 10-3/ (8.314 x 300) = 8.1 x 10-7 mole

Step II. To calculate the number of hydrogen molecules.

We know that

1 mol of hydrogen = 6.022 × 1023 molecules

8.1 x 10-7 mole = 6.022 × 1023 x 8.1 x 10-7 = 48.77 x 1016 molecules

Example 6

Calculate the mass of 120 mL of N2 at 150°C and 1 × 105 Pa pressure.

Solution: According to ideal gas equation,

pV = nRT =(m/M)RT

m = pVM/(RT)

p = 105 Pa , V = 120 ×10–6 m3

M = 28, T = 273 + 150 = 423 K,

R = 8.314 Nm K–1 mol–1

Mass of gas m = pVM/(RT) =(105 x 120 × 10–6 x 28)/(8.314 x 423) = 0.0955 g

Numerical Problems based on combined Gas law or Gas equation & ideal gas equation
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