Question

A mixture of nitrogen \(\left(\mathrm{N}_{2}\right)\) and helium has a volume of \(250 \mathrm{~mL}\) at \(30^{\circ} \mathrm{C}\) and a total pressure of \(745 \mathrm{mmHg}\). a. If the partial pressure of helium is \(32 \mathrm{mmHg}\), what is the partial pressure of the nitrogen? b. What is the volume of the nitrogen at STP?

Short answer

The partial pressure of nitrogen is 713 mmHg. The volume of nitrogen at STP is approximately 212.38 mL.

Step-by-step solution

Calculate Partial Pressure of Nitrogen

First, understand that the total pressure of the gas mixture is the sum of the partial pressures of the individual gases. Given that the total pressure (\text{P\text{total}}) is 745 mmHg, and the partial pressure of helium (\text{P\text{He}}) is 32 mmHg, the partial pressure of nitrogen (\text{P\text{N}\text{2}}) can be found using the formula: \[ P_{total} = P_{N_2} + P_{He} \] Rearrange this to solve for \(P_{N_2}\): \[ P_{N_2} = P_{total} - P_{He} \] Substitute the given values: \[ P_{N_2} = 745\text{ mmHg} - 32\text{ mmHg} = 713 \text{ mmHg} \]

Convert Volume of Nitrogen to STP

Now convert the volume of nitrogen at the given conditions (\text{V}) to its volume at standard temperature and pressure (STP: 0°C and 1 atm). Use the ideal gas law equation rearranged as: \[ P_1 V_1 / T_1 = P_2 V_2 / T_2 \] Where: \(P_1\) = 713 mmHg (partial pressure of nitrogen), \(V_1\) = 250 mL (initial volume), \(T_1\) = 30°C = 303.15 K (converted to Kelvin), and STP conditions are: \(P_2\) = 760 mmHg, \(T_2\) = 273.15 K. Rearrange the ideal gas law to solve for \(V_2\): \[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \] Substitute the known values: \[ V_2 = \frac{713 \text{ mmHg} \times 250 \text{ mL} \times 273.15 \text{ K}}{760 \text{ mmHg} \times 303.15 \text{ K}} \] Calculate to find the volume at STP: \[ V_2 \approx 212.38 \text{ mL} \]

Key concepts

Partial Pressure

Partial pressure refers to the individual pressure exerted by a specific gas in a mixture of gases. It is important because it helps us isolate the effect of each gas in the mixture.

Partial pressure is calculated using Dalton's Law of Partial Pressures. Dalton’s Law states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas. In mathematical form:

\[P_{total} = P_{gas1} + P_{gas2} + ... + P_{gasN} \]

This can be rearranged to solve for any individual gas:

\[P_{gasN} = P_{total} - (P_{gas1} + P_{gas2} + ... + P_{gas(N-1)}) \]

As presented in the exercise, the partial pressure of nitrogen in a nitrogen-helium mixture can be found by subtracting the partial pressure of helium from the total pressure. Given that the total pressure is 745 mmHg and the partial pressure of helium is 32 mmHg, the partial pressure of nitrogen can be calculated as:

\[P_{N_2} = 745 \text{ mmHg} - 32 \text{ mmHg} = 713 \text{ mmHg} \]

This means that nitrogen contributes 713 mmHg to the total pressure of the gas mixture.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation that describes the relationship between pressure, volume, temperature, and the number of moles of a gas. It’s expressed as:

\[PV = nRT \]

Where:

• P is the pressure of the gas (in units like atm or mmHg)
• V is the volume of the gas
• n is the number of moles of the gas
• R is the ideal gas constant (0.0821 atm·L/(mol·K) or 62.36 mmHg·L/(mol·K))
• T is the temperature in Kelvin

This law allows us to predict how a gas will behave under different conditions. For example, if we know the volume, pressure, and temperature of a gas, we can calculate how many moles of this gas are present.

To convert the volume of nitrogen at given conditions to its volume at STP using the Ideal Gas Law, we can rearrange it to:

\[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

We use this formula in the solution to find the new volume (\[V_2\]), given the initial conditions and STP conditions (0°C and 1 atm). By substituting the known values:

\[V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

Substituting the given values, we get:

\[V_2 = \frac{713 \text{ mmHg} \times 250 \text{ mL} \times 273.15 \text{ K}}{760 \text{ mmHg} \times 303.15 \text{ K}} \approx 212.38 \text{ mL} \]

This calculation shows that the volume of nitrogen at STP conditions is approximately 212.38 mL.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure (STP) is a set of conditions used as a reference point in gas law calculations. It provides a common framework to compare different gas behaviors. STP is defined as:

• Temperature: 0°C (273.15 K)
• Pressure: 1 atm (760 mmHg)

STP is particularly useful for standardizing comparisons and simplifying calculations. Whenever we need to find gas volumes under standard conditions, we use STP values to adjust the measured conditions to these standard settings.

In the example provided, we needed to convert the volume of nitrogen from given experimental conditions (250 mL at 30°C and 713 mmHg) to its volume at STP. The equation used for this conversion, derived from the Ideal Gas Law, is:

\[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Here, the conditions are known:

\[P_1 = 713 \text{ mmHg} \]

\[V_1 = 250 \text{ mL} \]

\[T_1 = 303.15 \text{ K} (30°C in Kelvin) \]

\[P_2 = 760 \text{ mmHg} (STP pressure) \]

\[T_2 = 273.15 \text{ K} (STP temperature) \]

Substitute these into the rearranged Ideal Gas Law to find \[V_2\], the volume at STP:

\[V_2 = \frac{713 \text{ mmHg} \times 250 \text{ mL} \times 273.15 \text{ K}}{760 \text{ mmHg} \times 303.15 \text{ K}} \]

The resulting volume is approximately 212.38 mL at STP. Understanding STP helps to simplify gas law calculations and allows easy comparison of different gases under standardized conditions.