Question

In what ratio by mass, the gases CO and 2 butene(C4H8) be mixed in a vessel so that they cause same partial pressure

Answer 1

Partial pressure of CO= Partial pressure of C₄H₈

SO,

$\chi _{CO}P_{T}= \chi _{C_{4}H_{8}}P_{T}$  

$\frac{mass of CO}{molar mass of CO}= \frac{mass of C_{4}H_{8}}{molar mass of C_{4}H_{8}}$

Here  

$\chi _{CO}$ , represents mole fraction of CO

$\chi _{C₄H₈}$ , represents mole fraction of C₄H₈

while,

$P_{T}$ , represents total pressure

That is

$\frac{n_{CO}}{n_{CO}+{n_{C_{4}H_{8}}}}=  \frac{n_{C_{4}H_{8}}}{n_{CO}+{n_{C_{4}H_{8}}}}$

$n_{CO}= n_{C_{4}H_{8}}$

$\frac{mass of CO}{molar mass of CO}=\frac{mass of C_{4}H_{8}}{molar mass of C_{4}H_{8}}$

$\frac{mass of CO}{63.5}=\frac{mass of C_{4}H_{8}}{56.1}$

$\frac{mass of CO}{mass of C_{4}H_{8}}=\frac{28}{56}=\frac{1}{2}$

so, the gases CO and 2 butene(C4H8) can be mixed in a vessel so that they cause same partial pressure in the ratio.

$\frac{1}{2}$

Answer 2

2-butene 56 g/mol

CO = 28 g/mol

for equal partial pressure they have to have the same mole concentration

Ptotal = Pp x Xa ; the same concentration in %

0.5 mol and 0.5 mol; 50% and 50 %

it means:

0.5 x 28 = 14 g; 0.5 x 56 = 28 g

14/ 28 = 1/2 ratio

I hope this helps you understand, if there is any confusion please leave the comment below!