Find #q#, #w#, #DeltaU#, #DeltaH#, and #DeltaS_(sys)#, #DeltaS_(surr)#, and #DeltaS_(t ot)# for #"1 mol"# of an ideal gas undergoing the following changes? The starting pressure is #"3.00 atm"# and ending pressure is #"1.00 atm"# at #27^@ "C"#.

DISCLAIMER: LONG ANSWER!

#(a)# REVERSIBLE ISOTHERMAL EXPANSION

For a reversible isothermal expansion, here are some key principles you should remember:

• For an ideal gas, #DeltaU = DeltaH = 0# when the temperature is constant (i.e. for isothermal processes). If you want, I can show you why. It's actually a rather neat proof that is a reasonable question to ask on an exam.
• Expansion of the system/gas gives you negatively-signed work with respect to the system.
• #deltaq = TdS# ONLY for a reversible process. Otherwise, it is less than or equal to it.

That means we know that #color(blue)(DeltaU = DeltaH = 0)# right off the bat. Now recall the
first law of thermodynamics with constant temperature:

#cancel(DeltaU)^(0) = q_(rev) + w_(rev) = q_(rev) - intPdV#

where the integral is possible because the work is reversible (infinitesimally slow). Remember that the #P# in work is external pressure.

As a result:

#q_(rev) = int PdV#

#= int_(V_1)^(V_2) (nRT)/VdV# #" "#(for ideal gases, #PV = nRT#)

#= nRTln(V_2/V_1)#

At constant temperature and mols of ideal gas, #P_1V_1 = P_2V_2#. So, #V_2/V_1 = P_1/P_2#. Therefore:

#color(blue)(q_(rev)) = -nRTln(P_2/P_1)#

#= -"1 mol" cdot R cdot (27 + "273.15 K")ln("1.00 atm"/"3.00 atm")#

#= 329.7R" J"#

#~~# #color(blue)("2741.7 J")#

From above, #color(blue)(w_(rev)) = -q_(rev) = color(blue)(-"2741.7 J")#. From the relation of entropy to reversible heat flow, we have:

#dS_(sys) = (deltaq_(rev))/T#

We define #q# for the system anyways, and so:

#color(blue)(DeltaS_(sys)) = int dS_(sys) = int 1/Tdeltaq_(rev)#

#= "2741.7 J"/("300.15 K") = color(blue)("9.134 J/K")#

I could not have done the integral of #C_P/T#, because it would be based on #((delS)/(delT))_P#, which is going to integrate to give zero contribution to the change in entropy at constant temperature.

However, the only contribution to the entropy for this reversible process is #((delS)/(delP))_T#, i.e. the change in entropy due to a isothermal change in pressure. So, we can check our answer by doing it another way.

#DeltaS = int_(P_1)^(P_2) ((delS)/(delP))_TdP#

From the Gibbs' free energy Maxwell Relation, #((delS)/(delP))_T = -((delV)/(delT))_P#, so we could have also gotten #DeltaS_(sys)# this way:

#color(blue)(DeltaS_(sys)) = -int_(P_1)^(P_2) (nR)/PdP#

#= -nRln(P_2/P_1)#

#= -"1.00 mol" cdot "8.314472 J/mol"cdot"K" cdot ln("1.00 atm"/"3.00 atm")#

#=# #color(blue)("9.134 J/K")# as before!

For a reversible process, the change in entropy of the system is opposite in sign and same magnitude as the surroundings by conservation of energy.

#color(blue)(DeltaS_(surr)) = -DeltaS_(sys) = color(blue)(-"9.134 J/K")#

And thus, for the universe,

#color(blue)(DeltaS_(t o t)) = DeltaS_(sys) + DeltaS_(surr) = color(blue)("0 J/K")#

#(b)# ISOTHERMAL EXPANSION AGAINST EXTERNAL PRESSURE OF 1.00 ATM

If we have a constant external pressure of #"1.00 atm"#, we know that #P# for the work is #"1.00 atm"#.

Since this is not necessarily reversible, we cannot integrate and simply have to follow the given path (as work and heat flow are path functions) and say the work is irreversible:

#w_(irr) = -PDeltaV#

From the ideal gas law, we need to then solve for the initial and final volumes separately.

#V_1 = (nRT)/(P_1) = ("1 mol" cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot "300.15 K")/("3.00 atm")#

#=# #"8.210 L"#

#V_2 = (nRT)/(P_2) = ("1 mol" cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot "300.15 K")/("1.00 atm")#

#=# #"24.63 L"#

As a result, we can then calculate the work:

#color(blue)(w_(irr)) = -1.00 cancel"atm" xx (24.63cancel"L" - 8.210cancel"L") xx "8.314472 J"/(0.082057 cancel("L"cdot"atm"))#

#= color(blue)(-"1664 J")#

As before, #DeltaU = DeltaH = 0#, so we can calculate #q# as:

#color(blue)(q_(irr)) = -w_(irr) = color(blue)("1664 J")#

For the change in entropy, this is trickier. I don't think we can calculate an exact value, since the process is not necessarily reversible, and we can't really assume it is "quasi-static".

[We can only use Maxwell Relations for reversible, or quasi-static irreversible processes in thermodynamically-closed systems.]

We actually have that:

#dS >= (deltaq)/T#

And since the process is not necessarily reversible, we say that

#dS_(sys) > (deltaq_(irr))/T#

So we can only (safely) give an inequality.

#DeltaS_(sys) > q_(irr)/T#

#> "1664 J"/"300.15 K"#

Therefore:

#color(blue)(DeltaS_(sys) > "5.544 J/K")#

And so,

#color(blue)(DeltaS_(surr) < -"5.544 J/K")#

This gives us an infinite number of possibilities for #DeltaS_(t ot)#. All we can say then is that by the second law of thermodynamics:

#color(blue)(DeltaS_(t ot) > "0 J/K")#