Question

2 L flask contains 1.6 g of CH4 ( methane) & 0.5 g of hydrogen at 27°C. Calculate the partial pressure of each gas in the mixture and calculate the total pressure........plzzzzzzzz help ....plzzzzz

Answer 1

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Answer 2

Answer : The partial pressure of methane and hydrogen are 1.23 atm and 3.07 atm respectively and total pressure is, 4.3 atm.

Solution :

First we have to calculate the partial pressure of methane and hydrogen gas.

$p_{CH_4}=\frac{w_{CH_4}\times R\times T}{M_{CH_4}\times V}$

$p_{CH_4}$ = partial pressure of methane

$w_{CH_4}$ = mass of methane = 1.6 g

$M_{CH_4}$ = molar mass of methane = 16 g/mole

T = temperature = $27^oC=273+27=300K$

R = gas constant = 0.0821 Latm/moleK

V = volume = 2 L

Now put all the values in the above formula, we get

$p_{CH_4}=\frac{1.6g\times 0.0821Latm/moleK\times 300K}{16g/mole\times 2L}=1.23atm$

$p_{H_2}=\frac{w_{H_2}\times R\times T}{M_{H_2}\times V}$

where,

$p_{H_2}$ = partial pressure of hydrogen

$w_{H_2}$ = mass of hydrogen = 0.5 g

$M_{H_2}$ = molar mass of hydrogen = 2 g/mole

$p_{H_2}=\frac{0.5g\times 0.0821Latm/moleK\times 300K}{2g/mole\times 2L}=3.07atm$

Now we gave to calculate the total pressure.

$P=p_{CH_4}+p_{H_2}=1.23+3.07=4.3atm$

Therefore, the partial pressure of methane and hydrogen are 1.23 atm and 3.07 atm respectively and total pressure is, 4.3 atm.