1.17 Activity

1/10 Activity

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 w_rev=-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 = nC_p ΔT.  We do not need to use C_p here because ΔT = 0 (it is isothermal, meaning temperature does not change).  Therefore, ΔH = 0.

Welcome to Chem 152!

I will post activities here that we do in quiz section.  I will also post any helpful information I find here.  For a review of activities from Chem 142, go to www.chem142.blogspot.com.