# Question #51c75

##### 1 Answer

#### Explanation:

All you have to do here is to use the **ideal gas law equation**, which looks like this

#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

Rearrange the equal to solve for

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

Now, before plugging in the values, make sure that he **units** you have for volume, pressure, and temperature **match** the units used in the expression of the universal gas constant.

#ul(color(white)(aaaacolor(black)("What you have")aaaaaaaaaacolor(black)("What you need")aaaaa))#

#color(white)(aaaaaacolor(black)("liters " ["L"])aaaaaaaaaaaaaaacolor(black)("liters " ["L"])aaaa)color(darkgreen)(sqrt())#

#color(white)(aaacolor(black)("atmospheres " ["atm"])aaaaaacolor(black)("atmospheres " ["atm"])aaa)color(darkgreen)(sqrt())#

#color(white)(aacolor(black)("degrees Celsius " [""^@"C"])aaaaaaaaacolor(black)("Kelvin " ["K"])aaaa)color(red)(xx)#

Notice that you must convert the temperature from *degrees Celsius* to *Kelvin*. To do that, use the following conversion factor

#color(blue)(ul(color(black)(T["K"] = t[""^@"C"] + 273.15)))#

You will have

#T = 44.0^@"C" - 273.15 = "317.15 K"#

Now you're ready to solve for

#n = (4.70 color(red)(cancel(color(black)("atm"))) * 34.0 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")))) * 317.15color(red)(cancel(color(black)("K"))))#

#color(darkgreen)(ul(color(black)(n = "6.14 moles")))#

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