Numerical Problems based on combined Gas law or Gas equation & ideal gas equation
Last updated on February 24th, 2022 at 08:21 pm
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 × 10–6 m3
T = 27 + 273 = 300 K,
R = 8.314 J mol–1K–1
Now, n = pV/(RT) = 101 x 10^3 x 500 × 10–6 / ( 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
