Question

a nitrogen gas from a 24.0 L container with the pressure of 2atm and an oxygen gas from a 12.0 L with a pressure of 2atm and 273K were mixed together in a 10.0 L container. what are the partial pressure exerted by each gas in the mixture and what is the total pressure? use ideal gas formula and mole fraction.

Answer 1

A nitrogen gas from a 24.0 L container with the pressure of 2 atm and an oxygen gas from a 12.0 L with a pressure of 2 atm and 273 K were mixed together in a 10.0 L container. What are the partial pressure exerted by each gas in the mixture and what is the total pressure? use ideal gas formula and mole fraction.

Solution:

$$\begin{array}{l} \text{for } N_2: \quad pN_2 \cdot V_{N_2} = n_{N_2} \cdot RT \Rightarrow \\ \quad n = \frac{2 \text{ atm} \times 24.0 \, \text{L}}{0.082057 \left( \frac{ \text{L} \cdot \text{ atm} }{ \text{moles} \cdot \text{K} } \right) \times 273 \, \text{K}} = \\ \quad = 2.143 \, \text{moles}; \end{array}$$ $$\text{for } O_2: \quad n = \frac{2 \text{ atm} \times 12.0 \, \text{L}}{0.082057 \left( \frac{ \text{L} \cdot \text{ atm} }{ \text{moles} \cdot \text{K} } \right) \cdot 273 \, \text{K}} = 1.071 \, \text{moles}.$$

A mixture of ideal gases is formed after mixing of $N_2$ and $O_2$:

The partial pressure:

$$p_i(N_2) = \frac{2.143 \, \text{moles} \cdot 0.082057 \, \text{L} \cdot \text{atm} \, \text{moles} \cdot \text{K}}{10 \, \text{L}} \times 273 \, \text{K} = 4.801 \, \text{atm}$$ $$p_i(O_2) = \frac{1.071 \, \text{moles} \cdot 0.082057 \, \text{L} \cdot \text{atm} \, \text{moles} \cdot \text{K}}{10.0 \, \text{L}} \times 273 \, \text{K} = 2.399 \, \text{atm}$$

Due to Dalton's Law:

$$P_{\text{total}} = p_i(N_2) + p_i(O_2) = 7.2 \, \text{atm}$$

Answer: $p_i(N_2) = 4.801 \, \text{atm}$

$p_i(O_2) = 2.399 \, \text{atm}$

$P_{\text{total}} = 7.200 \, \text{atm}$

Answer provided by www.AssignmentExpert.com