159 mL of hydrogen gas, measured at 25°C and 2.5 atm pressure, and 260 mL of nitrogen gas, measured at 25°C and 2.7 atm pressure, were forced into a 159 mL container at 25°C. What would be the pressure (in atm) of the mixture of gases now in the 159 mL container? Enter to 2 decimal places.

Question

23-10-2024
Chemistry

High School

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159 mL of hydrogen gas, measured at 25°C and 2.5 atm pressure, and 260 mL of nitrogen gas, measured at 25°C and 2.7 atm pressure, were forced into a 159 mL container at 25°C. What would be the pressure (in atm) of the mixture of gases now in the 159 mL container? Enter to 2 decimal places.

Answer :

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AshishSV AshishSV · Answer 1 · 2024-10-14T01:38:45-04:00

The pressure of the mixture of gases in the 159 ml container at 25°C is approximately 7.02 atm.

Given data:

For hydrogen gas: V = 159 ml = 0.159 L, P = 2.5 atm, T = 25°C = 298K

For nitrogen gas: V = 260 ml = 0.26 L, P = 2.7 atm, T = 25°C = 298K

The ideal gas constant, R = 0.0821 L·atm/mol·K

Using the formula:
[tex]n = \frac{PV}{RT}[/tex]

For hydrogen gas:
[tex]n_{H2} = \frac{2.5 \times 0.159}{0.0821 \times 298} \approx 0.0164 \, \text{mol}[/tex]

For nitrogen gas:
[tex]n_{N2} = \frac{2.7 \times 0.26}{0.0821 \times 298} \approx 0.0288 \, \text{mol}[/tex]

Calculate the total moles of gas in the mixture:[tex]n_{total} = n_{H2} + n_{N2} = 0.0164 \, \text{mol} + 0.0288 \, \text{mol} = 0.0452 \, \text{mol}[/tex]

Use the Ideal Gas Law to find the total pressure in the 159 ml (0.159 L) container:

The volume of the container is kept fixed, V = 0.159 L

From the Ideal Gas Law:
[tex]P_{total} = \frac{n_{total}RT}{V}[/tex]

[tex]P_{total} = \frac{0.0452 \, \text{mol} \times 0.0821 \, \text{Latm/molK} \times 298 \, \text{K}}{0.159 \, \text{L}} \approx 7.02 \, \text{atm}[/tex]