1/10/12 Activity – Solutions will be posted on blog at chem152B.blogspot.com
1. Argon is compressed adiabatically from an initial volume of 5 L to a final volume of 1 L against a pressure of 10 atm. Initial temperature is 306K. Find q, w, delta H, and delta E for this process. Do the signs of q and w make sense?
To Find w, use w=-PΔV. ΔV = 1L – 5L = -4L. P = 10 atm. w = 40 L*atm. Be sure to convert to units of Joules! 1 L*atm = 101.3 J. w = 4052 J (w is + for compressions.)
q = 0 because this process is adiabatic. By definition, heat transfer, or q, = 0. ΔE = q + w. ΔE = 4052 J
H = E + PV. Therefore, ΔH = ΔE + Δ(PV). Δ(PV) must be expanded using the product rule (as in calculus): Δ(PV) = ΔP*V + P*ΔV. The first term is zero because pressure is not changing. The second term = -w. Therefore, ΔH = Δ E –w = 0. ΔH = 0
2. One mole of argon is compressed from an initial volume of 5 L to a final volume of 1 L against an external pressure and is given sufficient time to thermally equilibrate with its surroundings at 306 K (it is an isothermal process). Find q, w, delta H, and delta E for this process. Do the signs of q and w make sense? Use the equation wrev=-nRT*ln(V2/V1) to solve for w.
By definition, an isothermal process is one where ΔE = 0. Therefore w = -q. Using the equation above for w, we get that w = 4095 J. (w is + for compressions, q is – for heat leaving the system.) ΔH = nCp ΔT. We do not need to use Cp here because ΔT = 0 (it is isothermal, meaning temperature does not change). Therefore, ΔH = 0.
In part two, why does an isothermal process (ΔT = 0) necessitate that ΔE = 0. Is it it because E = 3/2RT?
ReplyDeleteDanny,
Deletedelta(E)=nCv*delta(T) (always true) so when delta(T) is zero (i.e. the process is isothermal), delta(E) is also zero.
Katherine
Thanks. We have labs on exam weeks, yes?
ReplyDeleteAlso, whenever you post I imagine that somewhere in Staples Center, Kobe is blogging about general chemistry.
Danny
Danny,
ReplyDeleteWe do have labs this week. The lab schedule can be found on the syllabus. Also, little known fact: Kobe is really good at thermodynamics!
Katherine