Problem 69 - Chemistry Central Science

Question

A quantity of \(\mathrm{N}_{2}\) gas originally held at \(4.75\) atm pressure in a 1.00-L container at \(26^{\circ} \mathrm{C}\) is transferred to a 10.0-L container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(5.25\) atm and \(26^{\circ} \mathrm{C}\) in a 5.00-L container is transferred to this same container. What is the total pressure in the new container?

Step-by-Step Solution

Answer

The total pressure in the new container after transferring \(N_2\) and \(O_2\) gases to it is approximately \(3.21 \, atm\).

Step 1: Ideal Gas Law

The Ideal Gas Law is defined as: \(PV = nRT\) where: \(P\) = pressure \(V\) = volume \(n\) = number of moles of gas \(R\) = universal gas constant \(T\) = temperature in kelvins

Step 2: Convert temperatures to Kelvin

Convert given temperatures from Celsius to Kelvin using the following formula: \(K = °C + 273.15\) For N2 gas: \(T_1 = 26^{\circ} C + 273.15 = 299.15K\) and for the final container: \(T_2 = 20^{\circ} C + 273.15 = 293.15K\)

Step 3: Calculate the number of moles (n) for the N2 gas

Rearrange the Ideal Gas Law to find n: \(n = \frac{PV}{RT}\) For \(N_2\) gas, we have: \(P = 4.75 \, atm\) \(V = 1.00\, L\) \(T = 299.15\, K\) We will use the universal gas constant R with a value of 0.0821 L atm / (K mol): \(n_1 = \frac{4.75 \times 1.00}{0.0821 \times 299.15} = 0.193122 \, moles\)

Step 4: Calculate the number of moles (n) for the O2 gas

For \(O_2\) gas, we have: \(P = 5.25\, atm\) \(V = 5.00\, L\) \(T = 299.15\, K\) \(n_2 = \frac{5.25 \times 5.00}{0.0821 \times 299.15} = 1.073152 \, moles\)

Step 5: Calculate the total number of moles

The total number of moles, \(n_t \), is the sum of moles of \(N_2\) and \(O_2\): \(n_t = n_1 + n_2 = 0.193122 + 1.073152 = 1.266274 \, moles\)

Step 6: Find the total pressure (P) in the new container

We are given the final volume and temperature of the container: \(V_f = 10.0\, L\) \(T_f = 293.15\, K\) Now, apply the Ideal Gas Law to find the total pressure: \(P_f = \frac{n_tRT_f}{V_f} = \frac{1.266274 \times 0.0821 \times 293.15}{10.0} = 3.213144\, atm\) Therefore, the total pressure in the new container is approximately \(3.21 \, atm\).