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Burka [1]
3 years ago
14

What volume (in liters) does 3.91 moles of nitrogen gas at 5.35 atm and 323 K occupy

Chemistry
1 answer:
ivanzaharov [21]3 years ago
4 0
Data Given:
Moles = n = 3.91 mol

Pressure = P = 5.35 atm

Temperature = T = 323 K

Volume = V = ?

Formula used: Ideal Gas Equation is used,

P V = n R T
Solving for V,
V = n R T / P
Putting Values,
V = (3.91 mol × 0.0825 atm.L.mol⁻¹.K⁻¹ × 323 K) ÷ 5.35 atm

V = 19.36 L
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Explanation:

Expression for the v_{rms} speed is as follows.

            v_{rms} = \sqrt{\frac{3kT}{M}}

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               v_{rms} = \sqrt{\frac{3kT}{M}}

                            = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.02 K}{0.031 kg/mol}}

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Hence, v_{rms} for oxygen atom is 498.5 m/s.

For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its v_{rms} speed as follows.

                v_{rms} = \sqrt{\frac{3kT}{M}}

                              = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.92 K}{0.028 kg/mol}}

                              = 524.5 m/s

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