A 3.0 mol gas sample occupies 6.0 liters at 25 °C. What is the pressure of the gas in kPa’s?

2 Answers
Dec 2, 2015

Answer:

The pressure will be quite high!

Explanation:

#P=(nRT)/V#

#=# #{(3.0*cancel(mol)*0.0821*cancel(L)*atm*cancel(K^(-1))cancel(mol)^(-1)xx298cancelK)/(6.0*cancelL)}xx101.3*kPa*(atm)^-1#

Is the pressure you get sensible? Would you a priori expect it to be greater or lesser than atmospheric pressure?

Dec 2, 2015

Answer:

The pressure will be #"1240 kPa"#

Explanation:

Use the ideal gas law with the equation #PV=nRT#, where #P# is pressure, #V# is volume, #n# is moles, #R# is the gas constant, and #T# is the Kelvin temperature.

Given/Known
#V="6.0 L"#
#n="3.0 mol"#
#R="8.3144598 L kPa K"^(-1) "mol"^(-1)"#
https://en.wikipedia.org/wiki/Gas_constant
#T="25"^"o""C"+273.15="298 K"#

Unknown
pressure, #P#

Equation
#PV=nRT#

Solution
Rearrange the equation to isolate #P# and solve.

#P=(nRT)/V#

#P=(3.0cancel"mol"xx8.3144598cancel"L" "kPa" cancel("K"^(-1)) cancel("mol"^(-1))xx298cancel"K")/(6.0cancel"L")="1240 kPa"# (rounded to three significant figures) http://academic.umf.maine.edu/magri/PUBLIC.acd/tools/SigFigsAndRounding.html