At `27^@C`, a cylinder of 20 L capacity contains three gases He, `O_2 " and " N_2`. Their masses are 0.502 g, 0.250 g and 1.00 g respectively. If all these gases behave ideally, calculate the partial pressure of each gas as well as the total pressure.

Explanation

Calculation of moles :
Moles of helium = `(0.502)/4 =0.125`
Moles of oxygen = `(0.250)/32 = 7.81xx10^(-3)`
Moles of nitrogen = `(1.00)/28 = 3.57xx10^(-2)`
Calculation of partial pressures :
`:. " " PV=nRT`
`:. " " P= (nRT)/V`
Given V=20L and T=27+273=300K
Hence,
`P_(H_e)=(0.125xx0.0821xx300)/20 = 0.154` atm
`P_(O_2) = (7.81xx10^(-3)xx0.0821xx300)/20 = 9.62xx10^(-3)` atm
`P_(N_2) = (3.57xx10^(-2)xx0.0821xx300)/20 = 0.0440` atm
Calculation of total pressure :
According to Dalton.s law of partial pressure,
`P_(mixture) = P_(He)+P_(O_2)+P_(N_2)=0.154+9.62xx10^(-3)+0.0440 = 0.208` atm