Question

a cylinder was filled with gaseous mixture containing CO and N2 having equal mass the ratio of their partial pressure in cylinder are

Answer 1

P′N2=PT×molefractionofN2

P′CO=PT×molefractionofCO

Hello friend,

P′COP′N2=mole fraction of N2 mole fraction of CO

=molesofN2molesofCO

=w28w28

=1:1

I hope the above solution answered your question.

Thank You

Answer 2

Answer: The ratio of partial pressure of $CO:N_2$ is 1 : 1

Explanation:

The partial pressure of a gas is given by Raoult's law, which is:

$p_A=p_T\times \chi_A$     ......(1)

where,

$p_A$ = partial pressure of substance A

$p_T$ = total pressure

$\chi_A$ = mole fraction of substance A

We are given:

$m_{CO}=m_{N_2}$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

And,

$n_A=\frac{m_A}{M_A}$

Mole fraction of carbon monoxide is given as:

$\chi_{CO}=\frac{\frac{m_{CO}}{M_{CO}}}{(\frac{m_{CO}}{M_{CO}}+\frac{m_{N_2}}{M_{N_2}})}$

Molar mass of CO = 28 g/mol

Molar mass of $N_2$ = 28 g/mol

Putting values in above equation, we get:

$\chi_{CO}=\frac{\frac{m_{CO}}{28}}{(\frac{m_{CO}}{28}+\frac{m_{N_2}}{28})}\\\\\\\chi_{CO}=\frac{\frac{m_{CO}}{28}}{\frac{2m_{CO}}{28}}\\\\\\\chi_{CO}=\frac{1}{2}=0.5$

To calculate the mole fraction of nitrogen molecules, we use the equation:

$\chi_{CO}+\chi_{N_2}=1\\\\\chi_{N_2}=0.5$

Taking the ratios of partial pressures of the gases by using equation 1, we get:

$\frac{p_{CO}}{p_{N_2}}=\frac{p_T\times 0.5}{p_T\times 0.5}\\\\\frac{p_{CO}}{p_{N_2}}=\frac{1}{1}$

Hence, the ratio of partial pressure of $CO:N_2$ is 1 : 1