Worked examples · Sheet 3

Equilibrium states of the ideal gas

Theory used: Ideal gas Equation of state of the ideal gas Boyle's law and Gay-Lussac's laws Temperature scales

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A cylinder of volume $20.0 L$ contains oxygen at a pressure of $1.50 \cdot 1 0^{6} Pa$ and a temperature of $20.0^{\circ} C$ . Treating oxygen as an ideal gas, compute the number of moles and the mass of the gas (molar mass $M = 32.0 g/mol$ ).

p V = n R T ⟹ n = \frac{p V}{R T}, m = n M

V = 20.0 \cdot 1 0^{- 3} m^{3}, T = (20.0 + 273.15) K = 293.15 K

n = \frac{1.50 \cdot 1 0 ^{6} \cdot 20.0 \cdot 1 0 ^{- 3}}{8.314 \cdot 293.15} mol = \frac{3.00 \cdot 1 0 ^{4}}{2437.3} mol \approx 12.3 mol

m = 12.3 mol \cdot 0.0320 kg/mol \approx 0.394 kg

The equation of state of the ideal gas relates the four quantities $p$ , $V$ , $n$ , $T$ ; three are known, therefore the fourth is obtained by isolating it. Before substituting, the data are converted to the International System: the volume in cubic meters, with $1 L = 1 0^{- 3} m^{3}$ , and the temperature in kelvins, because the equation contains $T$ and not a temperature difference. With these units the product $p V$ is in joules and $R T$ in joules per mole, therefore the ratio is in moles.

The mass is obtained by multiplying by the molar mass expressed in $kg/mol$ : the cylinder contains about $394 g$ of oxygen. As an order-of-magnitude check, at atmospheric pressure and $0^{\circ} C$ one mole occupies about $22 L$ ; here the pressure is about fifteen times atmospheric pressure and the cylinder contains about a dozen moles in $20 L$ , a consistent value.

Exercise 2

A rigid, sealed container holds air at $1.00 \cdot 1 0^{5} Pa$ and $27.0^{\circ} C$ . The container is heated up to $177^{\circ} C$ . Compute the final pressure, treating air as an ideal gas.

V, n constant ⟹ \frac{p _{i}}{T _{i}} = \frac{p _{f}}{T _{f}} ⟹ p_{f} = p_{i} \frac{T _{f}}{T _{i}}

T_{i} = 300.15 K, T_{f} = 450.15 K

p_{f} = 1.00 \cdot 1 0^{5} \cdot \frac{450.15}{300.15} Pa \approx 1.50 \cdot 1 0^{5} Pa

The container is rigid and sealed, therefore the volume and the number of moles do not change. Writing the equation of state in the two states and dividing side by side, $n$ , $R$ and $V$ cancel and what remains is the second law of Gay-Lussac, $\frac{p}{T} = constant$ . The temperatures must be in kelvins, because the law is a direct proportionality only on the absolute scale.

The final pressure is one and a half times the initial one, since the absolute temperature goes from about $300 K$ to about $450 K$ . Mistakenly using the temperatures in degrees Celsius would give the ratio $\frac{177}{27} \approx 6.6$ , an increase in pressure with no physical meaning for a heating of $150 K$ .

Exercise 3

A cylinder of volume $10.0 L$ contains nitrogen at $2.00 \cdot 1 0^{6} Pa$ and $300 K$ . After some use the pressure has dropped to $5.00 \cdot 1 0^{5} Pa$ and the temperature to $290 K$ . Compute the mass of nitrogen that has left the cylinder ( $M = 28.0 g/mol$ ), treating the gas as ideal.

n_{1} = \frac{p _{1} V}{R T _{1}}, n_{2} = \frac{p _{2} V}{R T _{2}}

Δ m = (n_{1} - n_{2}) M = \frac{M V}{R} (\frac{p _{1}}{T _{1}} - \frac{p _{2}}{T _{2}})

\frac{p _{1}}{T _{1}} = 6.667 \cdot 1 0^{3} Pa/K, \frac{p _{2}}{T _{2}} = 1.724 \cdot 1 0^{3} Pa/K

Δ m = \frac{0.0280 \cdot 10.0 \cdot 1 0 ^{- 3}}{8.314} \cdot (6.667 - 1.724) \cdot 1 0^{3} kg = 3.368 \cdot 1 0^{- 5} \cdot 4.943 \cdot 1 0^{3} kg \approx 0.166 kg

The volume of the cylinder is fixed, while the pressure, the temperature and the number of moles change. The equation of state, applied separately to the initial state and to the final one, gives the moles present at the two instants: $n_{1} \approx 8.02 mol$ and $n_{2} \approx 2.07 mol$ . The difference is the number of moles that have left, and multiplying by the molar mass gives the mass.

Factoring out $\frac{M V}{R}$ before substituting reduces the computation to a single difference. About $166 g$ of nitrogen has left the cylinder, that is, about three quarters of the initial content; the ratio of the pressures is one quarter, and the slight decrease in temperature leaves in the cylinder a little more than one quarter of the initial moles.

Exercise 4

Two containers, of volumes $V_{A} = 2.0 L$ and $V_{B} = 3.0 L$ , are connected by a thin tube closed by a valve. The first contains an ideal gas at $p_{A} = 3.0 \cdot 1 0^{5} Pa$ , the second the same gas at $p_{B} = 1.0 \cdot 1 0^{5} Pa$ , both at the same temperature. The valve is opened and equilibrium is reached, at unchanged temperature. Compute the final pressure.

n_{A} + n_{B} = n_{f} ⟹ \frac{p _{A} V _{A}}{R T} + \frac{p _{B} V _{B}}{R T} = \frac{p _{f} ( V _{A} + V _{B} )}{R T}

p_{f} = \frac{p _{A} V _{A} + p _{B} V _{B}}{V _{A} + V _{B}}

p_{f} = \frac{3.0 \cdot 1 0 ^{5} \cdot 2.0 + 1.0 \cdot 1 0 ^{5} \cdot 3.0}{2.0 + 3.0} Pa = \frac{9.0 \cdot 1 0 ^{5}}{5.0} Pa = 1.8 \cdot 1 0^{5} Pa

The gas does not leave the system of the two containers, therefore the total number of moles is conserved; after the opening the gas occupies the volume $V_{A} + V_{B}$ with a single pressure $p_{f}$ . Each number of moles is written with the equation of state. The temperature is the same in all the states, therefore $R T$ cancels and the final pressure is the average of the initial pressures weighted by the volumes.

In the ratio the volumes appear both in the numerator and in the denominator, therefore they can be left in liters. As a check, $1.8 \cdot 1 0^{5} Pa$ lies between the two initial pressures and is closer to that of the larger container.