# Question #651b6

##### 1 Answer

#### Answer:

#### Explanation:

**!!! LONG ANSWER !!**

The idea here is that you can use the **ideal gas law** equation to find the pressure in box **before** mixing the two gases in that box, and **Boyle's Law** to calculate the pressure of methane **if placed alone** in box

At that point, you can use **Dalton's Law of Partial Pressures** to find the *total pressure* in box

So, you know that you're dealing with two boxes of *different volumes* that contain *different numbers of moles* of gas.

So, the ideal gas law equation looks like this

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "# , where

*universal gas constant*, usually given as

**absolute temperature** of the gas

If you take **of both boxes**, you can say that

#P_A * V_A = n_A * RT" " " "color(orange)(("*")) -># forbox#"A"#

Now, you know that

#V_B = 2 xx V_A -># the volume of box#"B"# istwicethat of box#"A"#

and

#n_B = 2 xx n_A -># box#"B"# containstwice as manymoles of gas as box#"A"#

You can thus use the ideal gas law equation to write

#P_B * V_B = n_B * RT -># forbox#"B"#

But this is equivalent to

#P_B * (2V_A) = (2n_A) * RT#

As you can see, you can use equation

#P_B * color(red)(cancel(color(black)(2)))V_A = color(red)(cancel(color(black)(2)))n_A * RT#

#{(P_B * V_A = n_A * RT), (P_A * V_A = N_A * RT):}#

the pressure in box **before the two gases are mixed** will be **equal** to the pressure in box

#P_B = P_A = "434.0 torr"#

At this point, you can use **Boyle's Law** to find the pressure exerted by the gas in box **if placed alone** in box

As you know, volume and pressure have an **inverse relationship** when temperature and number of moles are kept constant.

#color(blue)(|bar(ul(color(white)(a/a)P_1V_1 = P_2V_2color(white)(a/a)|)))" "# , where

In your case, you have

#P_A * color(red)(cancel(color(black)(V_A))) = P_("A in box B") * (2color(red)(cancel(color(black)(V_A))))#

This will get you

#P_("A in box B") = P_A/2 = "434.0 torr"/2 = "217.0 torr"#

So, the moles of methane will exert a pressure of **alone** in box

According to *Dalton's Law of Partial Pressures*, the **total pressure** of a gaseous mixture will be equal to the sum of the **partial pressures** of each constituent of the mixture if *left alone* in the **same volume** at the **same temperature** as the mixture.

You will thus have

#P_"total" = P_B + P_("A in box B")#

This will get you

#P_"total" = "434.0 torr" + "217.0 torr" = color(green)(|bar(ul(color(white)(a/a)"651.0 torr"color(white)(a/a)|)))#

**ALTERNATIVE APPROACH**

Here's an alternative method to use to find the **total pressure** of the mixture.

According to *Dalton's Law of Partial Pressures*, the **partial pressure** of a gas that's part of a gaseous mixture depends on its **mole fraction** and on the **total pressure** of the mixture.

#color(blue)(|bar(ul(color(white)(a/a)P_"gas i" = chi_"gas i" xx P_"total"color(white)(a/a)|)))#

Here

Now, we've established that the gas in box **mole fraction** of gas

#chi_B = "number of moles of helium"/"total number of moles of gas"#

#chi_B = (8 color(red)(cancel(color(black)("moles"))))/((8 + 4)color(red)(cancel(color(black)("moles")))) = 2/3#

This means that you have

#P_B = chi_B xx P_"total"#

and therefore

#P_"total" = P_B /chi_B#

This once again gets you

#P_"total" = "434.0 torr"/(2/3) = color(green)(|bar(ul(color(white)(a/a)"651.0 torr"color(white)(a/a)|)))#