For a #"313 g"# sample of helium gas that cools by #41.6^@ "C"# due to being compressed by #"267 L"# at a constant external pressure of #"1 atm"#, if its molar heat capacity is #"20.8 J/mol"^@ "C"#, calculate #DeltaE#, #q#, #w#, and #DeltaH#?

Question

Sebastian Acosta · Accepted Answer

Yes, this will take a while, so please be patient. It would be very helpful to know your derivations in this case. Well, first of all, let's get the pressure in terms of #"bar"#, because #"1 L"cdot"bar" = "100 J"# (you'll see what I mean later).  #P = "1.01325 bar"# Now, since we need to solve for all three of these, it would seem that we should be familiar with the first law of thermodynamics: #\mathbf(DeltaE = q + w)# where #DeltaE# is the change in internal energy caused by heat flow #q# and work #w = -P int_(V_1)^(V_2) dV#. The next thing we should know is how work is defined. We know that the temperature decreased, and that it doesn't matter what #T_2# or #T_1# are (#E# is a state function).  We know that the gas was compressed, so from the perspective of the gas, work was done on the gas. That means the gas let something else work upon it. Thus, the work will turn out to be numerically positive (a negative #DeltaV# in #-PDeltaV# gives #w > 0#). Additionally, I suppose that since the pressure is constant, we can use a handy thermodynamics relation: #DeltaH = DeltaE + Delta(PV)# #= q + w + (PDeltaV + VDeltaP + DeltaPDeltaV)# #= q - cancel(PDeltaV) + cancel(PDeltaV) + cancel(VDeltaP + DeltaPDeltaV)^(DeltaP = 0)# #\mathbf(DeltaH = q_p)#  in #"J"#. Now what about that molar heat capacity #barC_P# that you were given? Well, there's another nice relationship with molar enthalpy #DeltabarH#: #((delbarH)/(delT))_P = barC_P# (The molar heat capacity is the partial derivative of the molar enthalpy with respect to temperature at constant pressure.) Therefore, under continual pressure: #dbarH = barC_PdT# #\mathbf(DeltabarH = int_(T_1)^(T_2) barC_P dT)# in #"J/mol"# (not #"J"#). Since we aren't given the equation, we have to assume that the heat capacity will remain essentially linear in this temperature range (you can probably look it up in the NIST database): #n_"He"DeltabarH = color(blue)(q_p = n_"He"barC_PDeltaT)# where #n_"He"# is the #"mol"#s of helium, and #T_2 < T_1#. Since molar enthalpy #DeltabarH# is in #"J/mol"# but #DeltaE# is in #"J"#, we had to convert everything to #"J"#. Note that #\mathbf(C_P)# is actually in #\mathbf("J/"^@"C")#, NOT #\mathbf("J/mol"cdot""^@ "C")#.  #\mathbf(barC_P ne C_P)# Now we need #w# and #DeltaE#, which at this point we pretty much have. Take the integral of the work and that's it. #w = -P int_(V_1)^(V_2) dV# #color(blue)(w = -P(V_2 - V_1))# in #"L"cdot"bar"#, which you can convert to #"J"# (Wikipedia the universal gas constant #R# in #("L"cdot"bar")/("mol"cdot"K")# and compare it in #"J"/("mol"cdot"K")#). And lastly: #color(blue)(DeltaE) = q_p + w# #= color(blue)(n_"He"barC_PDeltaT - P(V_2 - V_1))# where #V_2 < V_1#. You can probably tell by now that the units are, of course, #"J"#. Yes, that's all there is to it. You just need to use the numbers that you have been given. #barC_P = "20.8 J/mol"^@ "C"#
#P = "1.01325 bar"#
#DeltaT = -41.6^@ "C"#
#DeltaV = -"267 L"#
#n_"He" = "313" cancel"g He" xx "1 mol He"/("4.0026" cancel"g He") = ...#

Adam Arnold · Answer

undefined $$ \Delta E = q - w $$

$$ q = m \cdot C \cdot \Delta T $$

$$ w = -P \cdot \Delta V $$

$$ \Delta H = \Delta E + P \cdot \Delta V $$

$$ \Delta E = (313 \, \text{g}) \cdot (20.8 \, \text{J/mol} \cdot ^\circ \text{C}) \cdot (-41.6 \, ^\circ \text{C}) $$

$$ q = (313 \, \text{g}) \cdot (20.8 \, \text{J/mol} \cdot ^\circ \text{C}) \cdot (-41.6 \, ^\circ \text{C}) $$

$$ w = - (1 \, \text{atm}) \cdot (267 \, \text{L}) $$

$$ \Delta H = \Delta E + (1 \, \text{atm}) \cdot (267 \, \text{L}) $$