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)= 7.81 xx 10^(-3)`
Moles of Nitrogen `= (1.00)/(28)= 3.57 xx 10^(-2)`
Calculation of partial pressures:
PV= nRT
Therefore, `P= (nRT)/(V)`
Given `V= 20 L and T= 27^(@)C= 273 + 27= 300K`
Hence, `P_(H e)= (0.125 xx 0.0821 xx 300)/(20)= 0.154` atm.
`P_(O_(2))= (7.81 xx 10^(-3) xx 0.0821 xx 300)/(20)`
`=9.62 xx 10^(-3)` atm.
`P_(N_(2))= (3.57 xx 10^(-2) xx 0.0821 xx 300)/(20)`
= 0.440 atm
Calculation of total pressure: According to Dalton.s law of partial pressure,
`P_("mixture")= P_(H e)+ P_(O_(2)) + P_(N_(2))`
`P_("mixture") = 0.154 + 9.62 xx 10^(-3) + 0.0440` = 0.208 atm.