# How many moles of gas are contained in 1.25 L at 303 K and 1.10 atm?

##### 1 Answer

#### Answer:

#### Explanation:

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

Now, take a look at the units used in the expression of the universal gas constant. The units you have for volume, temperature, and pressure **must** match those used in the expression of

#color(white)(aaaaaaaaaaacolor(Red)("Need")aaaaaaaaaaaaaaacolor(blue)("Have")aaaaa)#

#color(white)(aaaaa)color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaa)#

#color(white)(aaaaaaacolor(black)("Liters, L")aaaaaaaaaaaaacolor(black)("Liters, L")aaaaaaaaaaacolor(green)(sqrt())#

#color(white)(aaaaaaacolor(black)("Kelvin, K")aaaaaaaaaaaacolor(black)("Kelvin, K")aaaaaaaaaacolor(green)(sqrt())#

#color(white)(aaaaaaacolor(black)("Atmospheres, atm")aaaaacolor(black)("Atmospheres, atm")aaaacolor(green)(sqrt())#

Rearrange the ideal gas law equation to solve for

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

Plug in your values to find

#n = (1.10 color(red)(cancel(color(black)("atm"))) * 1.25color(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")))) * 303color(red)(cancel(color(black)("K")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.0553 moles")color(white)(a/a)|)))#

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