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A 6.69 L cylinder contains 1.62 mol of gas A and 4.00 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.

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princessesther2011

Answer:

The partial pressure of gas A is 5.97 atm

The partial pressure of gas B is 14.73 atm

Explanation:

From the ideal gas equation

PV = nRT

P = nRT/V

n is total moles of gas mixture = 1.62 + 4 = 5.62 moles

R = 82.057×10^-3 L.atm/gmol.K

T = 27.3°C = 27.3+273 = 300.3 K

V = 6.69 L

P (total pressure in the cylinder) = 5.62×82.057×10^-3×300.3/6.69 = 20.7 atm

Partial pressure of gas A = 1.62/5.62 × 20.7 atm = 5.97 atm

Partial pressure of gas B = 4/5.62 × 20.7 atm = 14.73 atm

olasunkanmilesanmi

Answer:

a. 5.96atm

b. 14.73atm

Explanation:

From the ideal gas law, the relationship between the volume,mole,temperature and pressure is given as

PV=nRT

Where P=pressure, V=volume, T=temperature, n=mole R=gas constant

For gas A

V=6.69L,

T=27.3C= 300.3K,

n=1.62mol,

We can calculate the partial pressure as

[tex]P=\frac{nRT}{V}\\ P=\frac{1.62*0.08206*300.3}{6.69}\\ P=5.96atm[/tex]

For gas B

V=6.69L,

T=27.3C= 300.3K,

n=4.00mol,

[tex]P=\frac{nRT}{V}\\ P=\frac{4.00*0.08206*300.3}{6.69}\\ P=14.73atm[/tex]

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