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
1
Expert's answer
2020-05-15T10:22:46-0400

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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