Exercises

Exercise 1. Derive the value of the gas constant, R, where the pressure is in torr and volume is in mL.

The ideal gas law is PV = nRT.

Solve for R.

\(\displaystyle R\;=\;\frac{PV}{nT}\)
 
At STP:

P = 760 mmHg, V = 22410. mL, n = 1 mole, T = 273.15 K

\(\displaystyle R\;=\;\frac{PV}{nT}\;=\;\frac{760\;mmHg\times\;22410\;mL}{1\;mol\times\;273.15\;K}\;=\;6.24\times\;10^4\;\frac{mmHg⋅mL}{mol⋅K}\)

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Exercise 2. A sample of N₂ gas has a mass of 2.20 g and a volume of 0.45 L at a pressure of 5.95 atm. What is the temperature in °C?

First convert the g of N₂ to moles.

\(\displaystyle 2.20\;g\times\;\frac{1\;mol}{28.013\;g}\;=\;0.0785\;mol\;N_2\)

Solve the ideal gas law, PV = nRT for T

\(\displaystyle T\;=\;\frac{PV}{nR}\;=\;\frac{5.95\;atm\times\;0.45\;L}{0.0821\frac{L⋅atm}{mol⋅K}\times\;0.0785\;mol}\;=\;415\;K\)
 
415 K – 273 = 142°C

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Exercise 3. What is the density, g/L, of krypton gas at STP?

Solve the ideal gas law for n/V. Then convert to grams.

PV = nRT, and n/V = P/RT

\(\displaystyle \frac{n}{V}\;=\;\frac{1.00\;atm}{0.0821\frac{L⋅atm}{mol⋅K}\times\;273.15\;K}\;=\;0.0446\;\frac{mol}{L}\)
 
The molar mass of krypton gas is 83.80 g/mol

\(\displaystyle 0.0446\frac{mol}{L}\times\;\frac{83.80\;g}{1\;mol}\;=\;\mathbf{3.74\;\frac{g}{L}}\)

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Exercise 4. A helium gas cylinder has a volume of 4.60 L and a pressure of 150 atm at 23.0°C. How many balloons can be filled if each balloon has a volume of 1.35 L and a pressure of 1.20 atm at 23.0°C?

nRT is constant. Now, we use P₁V₁ = P₂V₂ and solve for V₂.

\(\displaystyle V_2\;=\;\frac{P_1V_1}{P_2}\;=\;\frac{150\;atm\times\;4.60\;L}{1.20\;atm}\;=\;575\;L\)
 
\(\displaystyle 575\;L\times\;\frac{1\;balloon}{1.35\;L}\;=\;\mathbf{426\;baloons}\)

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Exercise 5. What is the density, g/L, of NH₃ gas if the pressure is 750. mmHg at 22.4°C?

The molar mass of NH₃ is 17.0 g/mol.

\(\displaystyle d\;=\;\frac{PM_m}{RT}\)
 
P = \(750\;mmHg\;\times\frac{1.00\;atm}{760\;mmHg}\;=\;0.987\;atm\)
M_m = 17.0 g/mol
T = 22.44°C + 273.15 = 295.55 K
R = 0.0821 \(\frac{L⋅atm}{mol⋅K}\)

\(\displaystyle d\;=\;\frac{0.987\;atm\times\;17.0\frac{g}{mol}}{0.0821\frac{L⋅atm}{mol⋅K}\times\;295.55\;K}\;=\;\mathbf{0.69\frac{g}{L}}\)

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Exercise 6. A 129 g sample of a liquid was vaporized in a 250. mL flask at 122°C and 788 mmHg. What is the molecular mass of the substance?

\(\displaystyle M_m\;=\;\frac{mRT}{PV}\)

m = 129 g
T = 122°C + 273 = 395 K
P = 788 mmHg \(\times\;\frac{1\;atm}{760\;mmHg}\;=\;1.037\;atm\)
V = 250 mL = 0.250 L
R = 0.0821 \(\frac{L⋅atm}{mol⋅K}\)

\(\displaystyle M_m\;=\;\frac{129\;g\times\;0.0821\frac{L⋅atm}{mol⋅K}\times\;395\;K}{1.037\;atm\times\;0.250\;L}\;=\;\mathbf{1.61\times\;10^4\;\frac{g}{mol}}\)


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Exercise 7. A reaction should yield 5.86 g of O₂. Calculate the volume at 22.3°C and 0.989 atm.

Convert g of O₂ to moles. The molar mass of O₂ is 31.9988 g/mol

\(5.86\;g\frac{1\;mol}{31.9988\;g}\;=\;0.183\;mol\)
 
Use the ideal gas law, PV = nRT and solve for V.

\(\displaystyle V\;=\frac{nRT}{P}\;=\;\frac{0.183\;mol\times\;0.0821\frac{L⋅atm}{mol⋅K}\times\;295.45\;K}{0.989\;atm}\;=\;\mathbf{4.49\;L}\)

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Exercise 8. Calculate the density of chloroform, CHCl₃, at 86.5°C and 795 mmHg.

\(\displaystyle d\;=\;\frac{PM_m}{RT}\)
 
M_m of CHCl₃ = 119.38 g/mol
T = 86.5°C + 273.15 = 359.65 K
\(P\;=\;795\;mmHg\times\;\frac{1\;atm}{760\;atm}\;=\;1.046\;atm\)
R = 0.0821 \(\frac{L⋅atm}{mol⋅K}\)
 

\(\displaystyle d\;=\;\frac{1.046\;atm\times\;119.38\frac{g}{mol}}{0.0821\;\frac{L⋅atm}{mol⋅K}\times\;359.65\;K}\;=\;\mathbf{4.23\frac{g}{L}}\)

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Exercise 9. Calculate the number of grams of butane gas, C₄H₁₀, with a pressure of 0.902 atm, at a temperature of 65.5°C, and a volume of 2.6 L.

Use the ideal gas law, PV = nRT and solve for n.

\(\displaystyle n\;=\;\frac{PV}{RT}\)
 
P = 0.902 atm
V = 2.6 L
T = 65.5°C + 273.15 = 338.65 K

\(\displaystyle n\;=\;\frac{0.902\;atm\;\times 2.6\;L}{0.0821\frac{L⋅atm}{mol⋅K}\times\;338.65\;K}\;=\;0.0844\;mol\)
 
Covert 0.0844 moles of butane to g. The molar mass of butane is 58.12 g/mol.

\(\displaystyle 0.0844\;mol\times\;\frac{58.12\;g}{mol}\;=\;\mathbf{4.91\;g\;butane}\)

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Exercise 10. A mass of 248 g of Cl₂ is required for an experiment. What would the volume be at 38.5°C and 1862 mmHg?

Convert g of Cl₂ to moles. The molar mass of Cl₂ is 70.9 g/mol

\(248\;g\frac{1\;mol}{70.9\;g}\;=\;3.498\;mol\)
 
T = 38.5°C + 273.15 = 311.73 K
\( P\;=\frac{1\;atm}{760\;mmHg}\;=\;2.45\;atm\)

Use the ideal gas law, PV = nRT and solve for V.

\(\displaystyle V\;=\frac{nRT}{P}\;=\;\frac{3.498\;mol\times\;0.0821\frac{L⋅atm}{mol⋅K}\times\;311.73\;K}{2.45\;atm}\;=\;\mathbf{36.5\;L}\)

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