CHEM: Dalton\'s law states that the total pressure. P_total, of a mixture of gas
ID: 537667 • Letter: C
Question
CHEM:
Dalton's law states that the total pressure. P_total, of a mixture of gases in a container equals the sum of the pressures of each individual gas: P_total = P_1 + P_2 + P_3 + The partial pressure of the first component, P_1, is equal to the mole fraction of this component, X_1, times the total pressure of the mixture: P_1 = X_1 times P_total The mole fraction, X, represents the concentration of the component in the gas mixture, so x_1 = moles of component/total moles in mixture Three gases (8.00 g of methane, CH_4, 18.0 g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to the same 10.0-L container. At 23.0 degree C, the total pressure in the container is 5.50 atm. Calculate the partial pressure of each gas in the container. Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane. A gaseous mixture of O_2 and N_2 contains 37.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 345 mnHg? Express you answer numerically in millimeters of mercury.Explanation / Answer
Part -A
Moles of ch4 = 8/molar mass of ch4 = 8/16 =0.5
No.of moles of ethane = 18/30= 0.6
No.of moles of propane = ? Moles
PV = nRT to find the total number of moles present in the gas
P= 5.50 atm
V= 10L
T=296k ,n=? moles
R = 0.0821 Latm/k/mole
5.5*10 = n*0.0821*296
n=55/24.302 =2.263
total no.of moles = 0.5+0.6+no.of moles of propane = 2.263
no.of moles of propane = 1.163
mole fraction of ch4 = no.of moles of ch4/total number of moles = 0.5/2.263=0.22
mole fraction of ethane =0.6/2.263 = 0.265
mole fraction of propane = 1.163/2.263 =0.514
Partial pressure of ch4=mole fraction of ch4*total pressure
= 0.22*5.50= 1.21 atm
partial pressure of ethane = 0.265*5.50 = 1.46atm
Partial pressure of propane = 0.514*5.50 = 2.83 atm
Part B
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