Question

A cylinder with a movable piston records a volume of \(12.6 \mathrm{~L}\) when \(3.0 \mathrm{~mol}\) of oxygen is added. The gas in the cylinder has a pressure of 5.83 atm. The cylinder develops a leak and the volume of the gas is now recorded to be \(12.1 \mathrm{~L}\) at the same pressure. How many moles of oxygen are lost?

Short answer

Answer: Approximately 0.12 moles of oxygen are lost due to the leak.

Step-by-step solution

Identifying given values and constants

First, let's identify given values and constants for both the initial and final states: Initial state: - Volume: \(V_1 = 12.6\ L\) - Pressure: \(P_1 = 5.83\ atm\) - Amount of moles: \(n_1 = 3.0\ mol\) Final state: - Volume: \(V_2 = 12.1\ L\) - Pressure: \(P_2 = 5.83\ atm\) (same pressure) - Amount of moles: \(n_2\) (unknown) We also have the ideal gas constant, \(R = 0.0821\ \frac{L\cdot atm}{mol\cdot K}\). Note that we don't have the temperature, but it won't be necessary, as we will see in the next steps.

Applying the ideal gas law to both states

Next, we will apply the ideal gas law to both the initial and final states. Initial state: \(P_1V_1 = n_1RT_1\) Final state: \(P_2V_2 = n_2RT_2\)

Combining the equations

Since the pressure is the same for both states, we can write these two equations as: \(P_1V_1 = n_1RT_1\) - equation 1 \(P_2V_2 = n_2RT_2\) - equation 2 We can notice that since \(P_1 = P_2\), we can divide equation 1 by equation 2: \(\frac{n_1RT_1}{n_2RT_2} = \frac{V_1}{V_2}\) Since the temperature and the gas are the same in both the initial and final states, \(T_1=T_2\) and \(R_1=R_2\). As a result, both R's and both T's will cancel out: \(\frac{n_1}{n_2} = \frac{V_1}{V_2}\)

Finding moles in the final state

Now, we can find the unknown moles in the final state (\(n_2\)): \(n_2 = n_1 \cdot \frac{V_2}{V_1}\) \(n_2 = 3.0\ mol \cdot \frac{12.1\ L}{12.6\ L}\) \(n_2 \approx 2.88\ mol\) (rounded to two decimal points)

Calculating moles of oxygen lost

Finally, we can calculate how many moles of oxygen are lost: Moles of oxygen lost \(= n_1 - n_2\) Moles of oxygen lost \(= 3.0\ mol - 2.88\ mol\) Moles of oxygen lost \(\approx 0.12\ mol\) (rounded to two decimal points) So, approximately \(0.12\ mol\) of oxygen are lost due to the leak.