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

1 Answer
Apr 5, 2016

Answer:

#"0.278 atm"#

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

#P# - the pressure of the gas
#V# - the volume it occupies
#n# - the number of moles of gas
#R# - the universal gas constant, usually given as #0.0821("atm" * "L")/("mol" * "K")#
#T# - the 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 #P# without making any 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 three sig figs