In a mixture of N2 and CO2 gases, the partial pressure of CO2 is 1.25 atm. The total pressure of the mixture is 5 atm. The mole fraction of N2 in the mixture is 0.8.
Question
In a mixture of $\mathrm{N}{2}$ and $\mathrm{CO}{2}$ gases, the partial pressure of $\mathrm{CO}{2}$ is $1.25$ atm. The total pressure of the mixture is $5$ atm. The mole fraction of $\mathrm{N}{2}$ in the mixture is 0.8.
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Answer
To determine the mole fraction of nitrogen (( \mathrm{N}_2 )) in a gas mixture involving nitrogen and carbon dioxide (( \mathrm{CO}_2 )), we start by understanding the given conditions:
- Partial pressure of ( \mathrm{CO}_2 ): ( 1.25 ) atm
- Total pressure of the mixture: ( 5 ) atm
We know that the mole fraction of nitrogen is represented as ( x ). Since the total mole fraction of all components in a mixture equals 1, the mole fraction of ( \mathrm{CO}_2 ) will be ( 1 - x ).
Using the definition of partial pressure, which is the product of the mole fraction of a gas and the total pressure:
[ P_{\mathrm{CO}2} = (1 - x) \cdot P{\text{total}} ]
Given that ( P_{\mathrm{CO}2} = 1.25 ) atm and ( P{\text{total}} = 5 ) atm, we set up the equation:
[ 1.25 = (1 - x) \times 5 ]
Solving for ( x ):
[ \begin{align*} 1.25 &= 5 - 5x \ 5x &= 5 - 1.25 \ 5x &= 3.75 \ x &= \frac{3.75}{5} \ x &= 0.75 \end{align*} ]
Thus, the mole fraction of nitrogen (( \mathrm{N}_2 )) in the mixture is ( \mathbf{0.75} ).
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