What volume does 0.0250 mole #H_2# occupy at 0.821 atm pressure and 300 K?

2 Answers
Nov 20, 2015

Answer:

The volume is #"0.750 dm"^3#

Explanation:

Hydrogen behaves like an ideal gas.

#P*V=n*R*T#, where

#P = "the pressure" = 0.821 color(red)(cancel(color(black)("atm")))*(1.01325*10^5 "Pa")/(1 color(red)(cancel(color(black)("atm"))))#

#P = "83 190" color(red)(cancel(color(black)("Pa"))) * "1N/m"^2/(1 color(red)(cancel(color(black)("Pa")))) = "83 190 N/m"^2#

#T= "the temperature"#

#n = "number of moles"#

#R= "ideal gas constant" = 8.314 color(red)(cancel(color(black)("J")))//("K·mol") * "1 N·m"/(1 color(red)(cancel(color(black)("J")))) = "8.314 N·m/(K·mol)"#

#V = "the volume"#

#V=(n*R*T)/P#

#V=((0.250 color(red)(cancel(color(black)("mol"))) * 8.314 color(red)(cancel(color(black)("N")))*"m")//(color(red)(cancel(color(black)("K·mol")))) · 300 color(red)(cancel(color(black)("K"))))/("83 190" color(red)(cancel(color(black)("N")))//"m"^2)#

#V = 7.50*10^-4 " m"^3 = "0.750 dm"^3#

Nov 20, 2015

Answer:

The volume will be #"0.7 L"#.

Explanation:

Use the ideal gas law.

#PV=nRT#, where #n# is moles, and #R# is the gas constant.

Given/Known
#P="0.821 atm"#
#n="0.0250 mol"#
#R="0.082057338 L atm K"^(-1) "mol"^(-1)#
#T="300 K"#

Unknown

#V#

Equation

#PV=nRT#

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

#V=(nRT)/P#

#V=((0.0250cancel"mol"xx0.08205733" L" cancel"atm" cancel("K"^(-1)) cancel("mol"^(-1)) xx 300cancel"K"))/(0.821cancel"atm")="0.7 L"# (rounded to one significant figure due to 300 K)