Question #115982

There are 100g of CH4 and 100g of Cl2 gas in a 5 L flask at 35C. Calculate the partial pressure of each gas and the total pressure

Expert's answer

Assuming the ideal gas behavior of CH4 and Cl2, the partial pressure pp can be calculated using the ideal gas law:

p=nRTVp=\frac{nRT}{V} ,

where nn is the number of the moles of a gas, RR is the ideal gas constant, 8.314 m3·Pa/(mol·K) , TT is the temperature in kelvin, 35+273.15 = 308.15 K and VV is the volume of the system, 5 L or 5·10-3 m3.

The number of the moles of a gas nn is:

n=mMn = \frac{m}{M},

where mm is the mass and MM is the molar mass of a gas. The molar masses of CH4 and Cl2 are 16.04 g/mol and 70.91 g/mol, respectively. Therefore:

n(CH4)=100 g16.04 g/mol=6.23n(CH_4) = \frac{100 \text{ g}}{16.04 \text{ g/mol}} = 6.23 mol,

n(Cl2)=100 g70.91 g/mol=1.41n(Cl_2) = \frac{100\text{ g}}{70.91\text{ g/mol}} = 1.41 mol.

The partial pressures of CH4 and Cl2 are:

pCH4=6.238.314308.155103=3.19106p_{CH_4} = \frac{6.23·8.314·308.15}{5·10^{-3}} = 3.19·10^6 Pa

pCl2=1.418.314308.155103=7.22105p_{Cl_2} = \frac{1.41·8.314·308.15}{5·10^{-3}} = 7.22·10^5 Pa.

According to the Dalton's law of partial pressures, the total pressure is the sum of the partial pressures:

ptot=pCH4+pCl2=(3.19+0.722)106=3.91106p_{tot} = p_{CH_4} + p_{Cl_2} = (3.19 + 0.722)·10^6 = 3.91·10^6 Pa.

Answer: the partial pressures of CH4 and Cl2 are 3.19·106 Pa and 7.22·105 Pa, respectively. The total pressure of the system is 3.91·106 Pa.


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