Question

After \(0.600 \mathrm{~L}\) of \(\mathrm{Ar}\) at \(1.20 \mathrm{~atm}\) and \(227^{\circ} \mathrm{C}\) is mixed with \(0.200 \mathrm{~L}\) of \(\mathrm{O}_{2}\) at 501 torr and \(127^{\circ} \mathrm{C}\) in a \(400-\mathrm{mL}\) flask at \(27^{\circ} \mathrm{C}\) what is the pressure in the flask?

Short answer

The pressure in the flask is approximately 1.33 atm.

Step-by-step solution

- Convert all temperatures to Kelvin

To convert Celsius to Kelvin, use the formula: \[ T(K) = T(°C) + 273.15 \]For Argon gas: \[ T_{\text{Ar}} = 227 + 273.15 = 500.15 \text{ K} \]For Oxygen gas: \[ T_{\text{O}_{2}} = 127 + 273.15 = 400.15 \text{ K} \]For the flask: \[ T_{\text{flask}} = 27 + 273.15 = 300.15 \text{ K} \]

- Convert all pressures to the same unit (atm)

Argon's pressure is already in atm.For Oxygen's pressure in torr, convert using: \[ 1 \text{ atm} = 760 \text{ torr} \]\[ P_{\text{O}_{2}} = \frac{501}{760} \text{ atm} \approx 0.659 \text{ atm} \]

- Use Ideal Gas Law to find moles of each gas

The Ideal Gas Law is given by: \[ PV = nRT \]First, for Argon gas: \[ n_{\text{Ar}} = \frac{P_{\text{Ar}} V_{\text{Ar}}}{RT_{\text{Ar}}} = \frac{1.20 \text{ atm} \times 0.600 \text{ L}}{0.0821 \text{ L·atm·K}^{-1}\text{·mol}^{-1} \times 500.15 \text{ K}} \approx 0.0176 \text{ mol} \]Next, for Oxygen gas: \[ n_{\text{O}_{2}} = \frac{P_{\text{O}_{2}} V_{\text{O}_{2}}}{RT_{\text{O}_{2}}} = \frac{0.659 \text{ atm} \times 0.200 \text{ L}}{0.0821 \text{ L·atm·K}^{-1}\text{·mol}^{-1} \times 400.15 \text{ K}} \approx 0.0040 \text{ mol} \]

- Find total moles of gas and apply Ideal Gas Law to the flask

Total moles of gas:\[ n_{\text{total}} = n_{\text{Ar}} + n_{\text{O}_{2}} = 0.0176 \text{ mol} + 0.0040 \text{ mol} = 0.0216 \text{ mol} \]Now use the Ideal Gas Law to find the pressure in the flask:\[ P_{\text{flask}} = \frac{n_{\text{total}} RT_{\text{flask}}}{V_{\text{flask}}} \]Given that the flask volume is 400 mL or 0.400 L and R = 0.0821 L·atm·K⁻¹·mol⁻¹:\[ P_{\text{flask}} = \frac{0.0216 \text{ mol} \times 0.0821 \text{ L·atm·K}^{-1}\text{·mol}^{-1} \times 300.15 \text{ K}}{0.400 \text{ L}} \approx 1.33 \text{ atm} \]

Key concepts

Gas Pressure Calculation

Gas pressure is an essential concept in understanding how gases behave in different conditions. It is defined as the force exerted by the gas particles per unit area on the walls of its container. This force is due to the collisions of the gas particles with the walls.
In this exercise, we had a sample of Argon gas with an initial pressure of 1.20 atm and we needed to convert the pressure of Oxygen gas from torr to atm. Remember, the conversion factor between torr and atm is: \[ 1 \text{ atm} = 760 \text{ torr} \] So, for 501 torr, we converted it as follows: \[ P_{O_{2}} = \frac{501}{760} \text{ atm} \approx 0.659 \text{ atm} \]
Once all pressures are in the same unit, it makes the calculations more straightforward.
The calculated pressure in the flask at the end is the sum of the pressures contributed by both gases, assuming ideal behavior using the Ideal Gas Law formula.

Mole Calculations

Moles are a fundamental unit in chemistry, representing the amount of substance. In this exercise, we needed to find out how many moles of Argon and Oxygen gases were present.
We used the Ideal Gas Law equation, \[ PV = nRT \] to calculate the moles for each gas. Here, \[ P \text{ (pressure)}, V \text{ (volume)}, n \text{ (number of moles)}, R \text{ (universal gas constant)}, \text{ and } T \text{ (temperature in Kelvin)} \]
For Argon: \[ n_{Ar} = \frac{P_{Ar} V_{Ar}}{RT_{Ar}} = \frac{1.20 \, \text{atm} \times 0.600 \, \text{L}}{0.0821 \, \text{L·atm·K}^{-1} \text{·mol}^{-1} \times 500.15 \, \text{K}} \approx 0.0176 \, \text{mol}\] For Oxygen: \[ n_{O_{2}} = \frac{P_{O_{2}} V_{O_{2}}}{RT_{O_{2}}} = \frac{0.659 \, \text{atm} \times 0.200 \, \text{L}}{0.0821 \, \text{L·atm·K}^{-1} \text{·mol}^{-1} \times 400.15 \, \text{K}} \approx 0.0040 \, \text{mol} \]
Adding these, we get the total number of moles in the flask.

Temperature Conversion

Temperature is a crucial parameter when dealing with gases. It's essential to convert all temperatures to the same unit – Kelvin (K) in this case. The conversion formula from Celsius to Kelvin is straightforward:\br> \[T(K) = T(°C) + 273.15 \]For instance, converting 227°C for Argon: \[ T_{Ar} = 227 + 273.15 = 500.15 \text{ K}\] Similarly, for Oxygen at 127°C: \[ T_{O_{2}} = 127 + 273.15 = 400.15 \text{ K} \] And for the flask at 27°C: \[ T_{flask} = 27 + 273.15 = 300.15 \text{ K} \] This ensures that all our calculations align and conform to the requirements of the Ideal Gas Law, where temperature must be in Kelvin.