Calculating partial pressure in a gas mixture A 6.00 L tank at 4.13 °C is filled

Question

Calculating partial pressure in a gas mixture A 6.00 L tank at 4.13 °C is filled with 17.1 g of dinitrogen monoxide gas and 18.4 g of carbon dioxide gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Be sure your answers have the correct number of significant digits. 00-100 mole fraction: dinitrogen monoxide partial pressure mole fraction carbon dioxide partial pressure: atm Total pressure in tank: atm Explanation Check

Explanation / Answer

Molar mass of N2O,

MM = 2*MM(N) + 1*MM(O)

= 2*14.01 + 1*16.0

= 44.02 g/mol

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

n(N2O) = mass of N2O/molar mass of N2O

= 17.1/44.02

= 0.3885

n(CO2) = mass of CO2/molar mass of CO2

= 18.4/44.01

= 0.4181

n(N2O),n1 = 0.3885 mol

n(CO2),n2 = 0.4181 mol

Total number of mol = n1+n2

= 0.3885 + 0.4181

= 0.8065 mol

Now calculate the total pressure

Given:

V = 6.0 L

n = 0.8065 mol

T = 4.13 oC

= (4.13+273) K

= 277.13 K

use:

P * V = n*R*T

P * 6 L = 0.8065 mol* 0.0821 atm.L/mol.K * 277.13 K

P = 3.058 atm

1)

for N2O:

mole fraction = mol of N2O / total moles

= 0.3885 / 0.8065

= 0.482

partial pressure = mole fraction * total pressure

= 0.482 * 3.058 atm

= 1.47 atm

2)

for CO2:

mole fraction = 1- mole fraction of N2O

= 1 - 0.482

= 0.518

partial pressure = mole fraction * total pressure

= 0.518 * 3.058 atm

= 1.58 atm

3)

total pressure = sum of all partial pressures

= 1.47 atm + 1.58 atm

= 3.05 atm

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