# What is the pressure of 0.540 mol of an ideal gas at 35.5 L and 223 K?

##### 1 Answer

#### Explanation:

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

Before plugging in your values, make sure that the **units** you have for volume and temperature **match** those used in the expression of the universal gas constant.

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

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

#color(white)(aaaaaaacolor(black)("Liters, L")aaaaaaaaacolor(black)("Liters, L")aaaacolor(green)(sqrt())#

#color(white)(aaaaaaacolor(black)("Kelvin, K")aaaaaaaacolor(black)("Kelvin, K")aaaacolor(green)(sqrt())#

Since you already know the *number of moles* of gas present in your sample, you can proceed to solve the ideal gas law equation for *unit conversions*.

#color(purple)(|bar(ul(color(white)(a/a)color(black)(PV = nRT implies P = (nRT)/V)color(white)(a/a)|)))#

Plug in your values to get

#P = (0.540color(red)(cancel(color(black)("moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 223color(red)(cancel(color(black)("K"))))/(35.5color(red)(cancel(color(black)("L"))))#

#P = color(green)(|bar(ul(color(white)(a/a)"0.278 atm"color(white)(a/a)|))) -># rounded to threesig figs