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3 years ago

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A 5.55 L cylinder contains 1.21 mol of gas A and 4.93 mol of gas B, at a temperature of 27.3 °C. Calculate the partial pressure

of each gas in the cylinder. Assume ideal gas behavior.

1 answer:

melisa1 [442]3 years ago

7 0

Answer:

Pressure of gas A = 544.324 kPa

Pressure of gas B = 2217.784 kPa

Explanation:

Data provided in the question:

Total volume of the cylinder, V = 5.55 L = 0.00555 m³ [1 m³ = 1000 L]

Moles on gas A, $n_a$ = 1.21 mol

Moles on gas A, $n_b$ = 4.93 mol

Temperature, T = 27.3°C = 27.3 + 273 = 300.3 K

Now,

Pressure = $\frac{nRT}{V}$

here,

R is the ideal gas constant = 8.314 J/mol.K

Therefore,

Pressure of gas A = $\frac{n_aRT}{V}$

= $\frac{1.21\times8.314\times300.3}{0.00555}$

= 544324.32 Pa

= 544.324 kPa

Pressure of gas B = $\frac{n_bRT}{V}$

= $\frac{4.93\times8.314\times300.3}{0.00555}$

= 2217784.22 Pa

= 2217.784 kPa

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