For a constant-temperature, constant-volume process, what is the change in pressure caused by introducing #"3 mols"# of an additional ideal gas into a #"24-L"# contain at #25^@ "C"#?

#DeltaP = (DeltancdotRT)/V#

We have the following assumptions:

• The volume is constant (#DeltaV = 0#).
• The temperature is constant (#DeltaT = 0#).

We begin by assuming the gas is ideal and use the ideal gas law:

#PV = nRT#

where #P#, #V#, #n#, #R#, and #T# are the pressure, volume, mols, universal gas constant, and temperature, respectively (temperature must be in #"K"#).

Pressure, volume, mols, and temperature are all state functions, and we can write

#P_1V = n_1RT#,

#P_2V = n_2RT#,

since more mols of the same gas were added, but #V_1 = V_2 -= V# and #T_1 = T_2 -= T#.

Subtracting these equations gives:

#(P_2 - P_1)V = (n_2 - n_1)RT#

#DeltaPcdotV = DeltancdotRT#

Thus, the change in pressure is given by:

#color(blue)(DeltaP = (DeltancdotRT)/V)#

For example, if the pressure were in #"bar"#, #R = "0.083145 L"cdot"bar/mol"cdot"K"#, and volume is in #"L"#:

#DeltaP = ((Deltan " mols")("0.083145 L"cdot"bar/mol"cdot"K")(T" K"))/(V" L")#

If we had a change in mols of #"3 mols"# and a temperature of #"298.15 K"# and volume of #"24 L"#, then for an ideal gas:

#DeltaP = (("3 mols")("0.083145 L"cdot"bar/mol"cdot"K")("298.15 K"))/("24 L") = ???#