Question

there is a mixture of hydrogen helium and methane occupying a vessel of voloume 13 dm^3 at 37C and 1 atm pressure the mass of hydrogen and helium are 0.8g and 0.12g respectively . claculate the partial pressure of each gas in torr

Answer 1

Assuming the mixture to behave as an ideal gas.

molecular weight of CO2 = 12+16+16 = 44

molecular weight of SO2 = 32+16+16 = 64

moles of CO2 = 7.5/44 = 0.17045

moles of SO2 = 10/64 = 0.15625

n= moles of CO2 + moles of SO2 = 0.15625 + 0.17045 = 0.3267

V= 2.5L = 2.5*10^-3  m^3

T= 25C = 298K

PV = nRT

P = nRT/V = 0.3267*8.314*298/(2.5*10^-3) = 323769.10896 Pa

  = 3.1953atm

Answer 2

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Assuming the mixture to behave as an ideal gas.

molecular weight of CO2 = 12+16+16 = 44

molecular weight of SO2 = 32+16+16 = 64

moles of CO2 = 7.5/44 = 0.17045

moles of SO2 = 10/64 = 0.15625

n= moles of CO2 + moles of SO2 = 0.15625 + 0.17045 = 0.3267

V= 2.5L = 2.5*10^-3  m^3

T= 25C = 298K

PV = nRT

P = nRT/V = 0.3267*8.314*298/(2.5*10^-3) = 323769.10896 Pa

  = 3.1953atm