# Question #e9c20

##### 1 Answer

#### Explanation:

Your tool of choice here will be the **ideal gas law equation**

#color(blue)(ul(color(black)(PV = nRT)))#

Here

#P# is the pressure of the gas#V# is the volume it occupies#n# is the number of moles of gas present in the sample#R# is theuniversal gas constant, equal to#0.0821("atm L")/("mol K")# #T# is theabsolute temperatureof the gas

The first thing that you need to do here is to sue the ideal gas law equation to figure out how many moles of nitrogen gas are present in the initial sample.

#n_1 = (P_1 * V)/(R * T_1)#

#n_1 = (1.00 color(red)(cancel(color(black)("atm"))) * 1.00 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))#

#n_1 = "0.04085 moles N"_2#

Now, you know that the number of moles of nitrogen gas present in the sample is **increased** by the addition of the

#2.00 color(red)(cancel(color(black)("g"))) * "1 mole N"_2/(28.0134color(red)(cancel(color(black)("g")))) = "0.07139 moles N"_2#

Moreover, the temperature of the gas **decreases** from

#25^@"C" + 273.15 = "298.15 K"#

to

#-55^@"C" + 273.15 = "218.15 K"#

So, if you write the ideal gas law equation for the **initial state** of the gas as

#P_1 * V = n_1 * R * T_1#

you can write the ideal gas law equation for the **final state** of the gas as

#P_2 * V = n_2 * R * T_2#

Divide these two equations to get

#(P_1 * color(red)(cancel(color(black)(V))))/(P_2 * color(red)(cancel(color(black)(V)))) = (n_1 * color(red)(cancel(color(black)(R))) * T_1)/(n_2 * color(red)(cancel(color(black)(R))) * T_2)#

#P_1/P_2 = (n_1 * T_1)/(n_2 * T_2#

Rearrange to solve for

#P_2 = n_2/n_1 * T_2/T_1 * P_1#

Here you have

#n_2 = "0.04085 moles" + "0.07139 moles" = "0.11224 moles"#

Plug in your values to find

#P_2 = (0.11224 color(red)(cancel(color(black)("moles"))))/(0.04085color(red)(cancel(color(black)("moles")))) * (218.15 color(red)(cancel(color(black)("K"))))/(298.15color(red)(cancel(color(black)("K")))) * "1.00 atm"#

#color(darkgreen)(ul(color(black)(P_2 = "2.0 atm")))#

The answer is rounded to two **sig figs**, the number of sig figs you have for the two temperatures.