Question

A quantity of N2(g originally held at 3.80 atm in a 1.00L container at 26oC is transferred to a 10.0L container at 20oC. A quantity of O2(g is originally at 4.75 atm and 26oC in a 5.00L container is transferred into the same container. What is the TOTAL PRESSURE in the new container?

Answer 1

A quantity of N2(g originally held at 3.80 atm in a 1.00L container at 26oC is transferred to a 10.0L container at 20oC. A quantity of O2(g is originally at 4.75 atm and 26oC in a 5.00L container is transferred into the same container. What is the TOTAL PRESSURE in the new container?

**Solution.**

Find the quantity of $N_{2}$:

$$n(N_{2}) = \frac{P(N_{2}) * V(N_{2})}{R * T(N_{2})} = \frac{3.8 * 101325 * 1 * 10^{-3}}{8.314 * (273 + 26)} = 0.155 \text{ moles};$$

Find the quantity of $O_{2}$:

$$n(O_{2}) = \frac{P(O_{2}) * V(O_{2})}{R * T(O_{2})} = \frac{4.75 * 101325 * 5 * 10^{-3}}{8.314 * (273 + 26)} = 0.968 \text{ moles};$$

Find the total quantity of gases:

$$n_{\text{total}} = n(N_{2}) + n(O_{2}) = 0.155 + 0.968 = 1.123 \text{ moles};$$

Find the total pressure of gases in new container:

$$P_{\text{total}} = \frac{n_{\text{total}} * R * T_{\text{total}}}{V_{\text{total}}} = \frac{1.123 * 8.314 * (273 + 20)}{10 * 10^{-3}} = 273563 \text{ pascals} = \frac{273563}{101325} = 2.7 \text{ atm}.$$

Answer: the total pressure of gases is 2.7 atm.