# If the volume of the container is #"1.00 L"#, what is the change in mols of gas required to change the pressure from #"1.00 atm"# to #"0.920 atm"# if the temperature was also dropped from #25^@ "C"# to #22^@ "C"#? (#R = "0.082057 L"cdot"atm/mol"cdot"K"#)

##### 1 Answer

I get about

Since you are given:

#V = "1.00 L"# , thevolume#P_1 = "1.00 atm"# , theinitial pressure#P_2 = "0.920 atm"# , thefinalpressure#T_1 = 25^@ "C"# , theinitial temperature#T_2 = 22^@ "C"# , thefinaltemperature#R = "0.082057 L"cdot"atm/mol"cdot"K"# , theuniversal gas constant

we know that the volume had to have remained constant (we only got one value for it).

Also, since we were given pressure, temperature, volume, and

#PV = nRT#

Since we have the initial and final values for pressure and temperature, we have two variations on the ideal gas law:

#P_1V = n_1RT_1#

#P_2V = n_2RT_2#

So finding "the mols" should mean finding the **change in the mols**,

A reasonable way to do this is to solve for the initial

#n_1 = (P_1V)/(RT_1)#

#= (("1.00 atm")("1.00 L"))/(("0.082057 L"cdot"atm/mol"cdot"K")("25 + 273.15 K"))#

#=# #"0.0409 mols"#

So,

#n_2 = (P_2V)/(RT_2)#

#= (("0.920 atm")("1.00 L"))/(("0.082057 L"cdot"atm/mol"cdot"K")("22 + 273.15 K"))#

#=# #"0.0380 mols"#

So, if you wanted the change in

#color(blue)(Deltan)#

#=# #"0.0409 - 0.0380 mols"#

#=# #color(blue)("0.00289 mols")#