Question

Suppose that \(1.29 \mathrm{~g}\) of argon gas is confined to a volume of \(2.41 \mathrm{~L}\) at 29 C. What would be the pressure in the container? What would the pressure become if the temperature were raised to 42 " \(\mathrm{C}\) without a change in volume?

Short answer

The pressure of the argon gas at \(29 \,°C\) in the container is \(1.04 \, atm\). When the temperature is raised to \(42 \,°C\) without changing the volume, the pressure increases to \(1.07 \,atm\).

Step-by-step solution

Compute the Number of Moles

Use the molar mass of argon, which is 39.948 g/mol, to compute the number of moles (\(n\)) present in 1.29 g of argon. You can calculate \(n\) using the formula \[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{1.29 \, \text{g}}{39.948 \, \text{g/mol}} = 0.0323 \, \text{mol} \]

Convert Temperatures to Kelvin

The temperatures given are in degrees Celsius, and they need to be converted to Kelvin, since the Ideal Gas Law uses the Kelvin scale. The conversion is done using the formula \(T(K)=T(°C)+273.15\). Hence, the given temperatures become \[ T_1 = 29 \, °C + 273.15 = 302.15 \, K \] and \[ T_2 = 42 \, °C + 273.15 = 315.15 \, K \]

Compute the Initial Pressure

With \(n\), \(V\), and \(T\) known, we can now compute the pressure (\(P\)) using the Ideal Gas Law. Rearranging \(PV = nRT\) to solve for \(P\), we get \[ P = \frac{nRT}{V} \] Using \(R = 0.0821 \, \text{L atm/(mol K)}\), \(n = 0.0323 \, \text{mol}\), \(V = 2.41 \, \text{L}\), and \(T = 302.15 \, \text{K}\), we find \[ P = \frac{(0.0323 \, \text{mol})(0.0821 \, \text{L atm/(mol K)})(302.15 \, \text{K})}{2.41 \, \text{L}} = 1.04 \, \text{atm} \] So, the pressure of the gas at 29 °C is 1.04 atm.

Compute the Final Pressure

When the temperature is raised to 42 °C without a change in volume, the pressure can be found by using the ideal gas equation again, but this time with the new temperature \(T_2 = 315.15 \, \text{K}\). The new pressure \(P_2\) is given by \[ P_2 = \frac{nRT_2}{V} \] Substituting \(n = 0.0323 \, \text{mol}\), \(R = 0.0821 \, \text{L atm/(mol K)}\), \(T_2 = 315.15 \, \text{K}\), and \(V = 2.41 \, \text{L}\), we find \[ P_2 = \frac{(0.0323 \, \text{mol})(0.0821 \, \text{L atm/(mol K)})(315.15 \, \text{K})}{2.41 \, \text{L}} = 1.07 \, \text{atm} \] So, the pressure of the gas at 42 °C, keeping the volume constant, is 1.07 atm.

Key concepts

Molar mass of argon

Understanding the molar mass of argon is critical when working with gas calculations. It represents the mass of one mole of argon atoms. Precisely, the molar mass of argon is 39.948 grams per mole (g/mol). This value is derived from the atomic mass of argon, as listed on the periodic table, and is used to convert between grams and moles of the substance.

For instance, in the exercise, 1.29 grams of argon gas is present. To find the number of moles (n), we use the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}}\]
Substituting the given values:
\[ n = \frac{1.29 \, \text{g}}{39.948 \, \text{g/mol}}\]
This calculation allows us to determine the amount of argon in moles, which is essential for applying the Ideal Gas Law.

Converting Celsius to Kelvin

In thermodynamic gas equations, temperature needs to be in the Kelvin scale. The reason for this is that Kelvin is an absolute scale with its zero point at absolute zero, the theoretical point where particles have minimal vibrational motion.

To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature:
\[ T(K) = T(°C) + 273.15\]
The exercise provided temperatures of 29°C and 42°C. These convert to:
\[ T_1 = 29 \, °C + 273.15 = 302.15 \, K\]\[ T_2 = 42 \, °C + 273.15 = 315.15 \, K\]
These conversions ensure the temperatures are in the correct scale for use in the Ideal Gas Law equation.

Calculating gas pressure

To calculate the pressure of a given gas, you can apply the Ideal Gas Law, which is expressed as \(PV = nRT\). In this formula, \(P\) stands for pressure, \(V\) for volume, \(n\) for the moles of gas, \(R\) for the ideal gas constant (0.0821 L atm/(mol K)), and \(T\) for temperature in Kelvin.

When utilizing the Ideal Gas Law, it is crucial to ensure that all units are consistent. For example, if pressure is sought in atmospheres (atm), then the volume should be in liters (L), the molar volume in liters per mole per Kelvin (L mol-1 K-1), and the temperature in Kelvin (K). To solve for pressure, the equation is rearranged to:
\[ P = \frac{nRT}{V}\]
By simply substituting the known values of moles, temperature, and volume into this equation, one can determine the gas pressure at any given state, assuming ideal behavior where the gas particles do not interact and occupy no volume. This theoretical assumption allows for the straightforward calculation of pressure, as demonstrated in the step-by-step solution from the exercise.