# When an airbag is inflated the nitrogen gas has a pressure of 1.30 atmospheres, a temperature of 301 K, and a volume of 40.0 liters. What is the volume of the nitrogen at STP?

##### 1 Answer

#### Answer:

#### Explanation:

You can actually use two approaches to solve this problem.

You can use the **ideal gas law** to find the *number of moles* of gas present in that sample, then use the known **molar volume of a gas** at STP conditions to find the new volume.

As a way to double-check the result you get by using this approach, you can use the **combined gas law** equation to find the new volume of the gas *without* finding out how many moles it contains.

So, the ideal gas law equation looks like this

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

Here you have

#P# - the pressure of the gas

#V# - the volume it occupies

#n# - the number of moles of gas

#R# - theuniversal gas constant, usually given as#0.0821("atm" * "L")/("mol" * "K")#

#T# - theabsolute temperatureof the gas

Rearrange the equation to solve for

#PV = nRT implies n = (PV)/(RT)#

Plug in your values to find

#n = (1.30 color(red)(cancel(color(black)("atm"))) * 40.0 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")))) * 301color(red)(cancel(color(black)("K")))) = "2.104 moles N"_2#

Now, **STP** conditions are * currently* defined as a pressure of

**one mole**of any ideal gas occupies

**molar volume of a gas**at STP.

In your case, **moles** of nitrogen gas will occupy

#2.104 color(red)(cancel(color(black)("moles N"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole N"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("47.8 L")color(white)(a/a)|)))#

Now, you can double-check the result by suing the **combined gas law**

#color(blue)(|bar(ul(color(white)(a/a)(P_1V_1)/T_1 = (P_2V_2)/T_2color(white)(a/a)|)))#

Here

At mentioned before, **STP** conditions are defined as

#P_2 = "100 kPa" = 101/101.325color(white)(a)"atm"#

#T_2 = 0^@"C" + 273.15 = "273.15 K"#

Rearrange the combined gas law equation to solve for

#(P_1V_1)/T_1 = (P_2V_2)/T_2 implies V_2 = P_1/P_2 * T_2/T_1 * V_1#

Plug in your values to find

#V_2 = (1.30 color(red)(cancel(color(black)("atm"))))/(100/101.325color(red)(cancel(color(black)("atm")))) * (273.15color(red)(cancel(color(black)("K"))))/(301color(red)(cancel(color(black)("K")))) * "40.0 L"#

#V_2 = color(green)(|bar(ul(color(white)(a/a)color(black)("47.8 L")color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.

**SIDE NOTE** *STP conditions are often in accordance to their old definition of a pressure of*

*and a temperature of*

*Under these conditions, one mole of any ideal gas occupies*

*If this is the STP definition given to you, simply redo the calculations using*

*instead of*