# Question #e29ae

##### 1 Answer

#### Explanation:

The idea here is that you can use **Graham's Law of Effusion** to express the rate of effusion of hydrogen gas, *relative* to the rate of effusion of carbon dioxide,

As you know, **effusion** is the process in which the molecules of a gas escape through a pinhole in their container.

The *rate of effusion* essentially tells you the **number of molecules** of gas that escape through the pinhole per unit of time, usually *per second*.

According to **Graham's Law**, the rate at which a gas effuses is **Inversely proportional** to the square root of its **molar mass**

#color(blue)(|bar(ul(color(white)(a/a)"rate of effusion" prop 1/sqrt("molar mass")color(white)(a/a)|)))#

This means that **heavier** molecules will effuse *slower* than **lighter** molecules.

In your case, the rate of effusion of hydrogen gas relative to the rate of effusion of carbon dioxide can be expressed as

#"rate H"_ 2/"rate CO"_ 2 = 1/sqrt(M_("M H"_2)) * sqrt(M_("M CO"_2))#

#color(green)(|bar(ul(color(white)(a/a)color(black)("rate H"_ 2/"rate CO"_ 2 = sqrt( M_ ("M CO"_ 2)/M_ ("M H"_ 2)))color(white)(a/a)|)))#

Here

**molar mass** of carbon dioxide, equal to

**molar mass** of hydrogen gas, equal to

All you have to do now is plug in these values to find

#"rate H"_ 2/"rate CO"_ 2 = sqrt( (44.01 color(red)(cancel(color(black)("g mol"^(-1)))))/(2.016color(red)(cancel(color(black)("g mol"^(-1)))))) = color(green)(|bar(ul(color(white)(a/a)4.67color(white)(a/a)|)))#

This means that hydrogen gas will effuse **times faster** than carbon dioxide.

This means that in a *given period of time*, you can expect **times more molecules** of hydrogen gas to escape through the pinhole in their container than molecules of carbon dioxide.