1.31 Activity

1. A system does 3 J of work on the surroundings, and 12 J of work are added to the system.  What is the energy change of the system? 9 J What is the energy change of the surroundings? -9 J

2. A quantity of 0.850 mol of an ideal gas initially at a pressure of 15.0 atm and 300K is allowed to expand isothermally until its final pressure is 1.00 atm.  Calculate the value of the work done if the expansion is carried out (a) against a vaccum, (b) against a constant external pressure of 1.00 atm, and (c) reversibly – use the equation w_rev=-nRT*ln(P₁/P₂).

a.) w=-P_extDV, here P_ext = 0, so w=0

b.) P_ext=1.00 atm.  Find initial and final volumes using the ideal gas law:

      V₁=nRT/P₁; V₂=nRT/P₂; w=-P_extDV, where P_ext is equal to P₂

      DV= V₂-V₁ = nRT({1/P₂}-{1/P₁})

      Therefore, w=-nRTP₂({1/P₂}-{1/P₁}) = -19.5 L*atm = -1.98*10³ J

c.) for an isothermal, reversible process, we use the following equation for work:

     w_rev=-nRT*ln(V₂/V₁)=-nRT*ln(P₂/P₁) = -5.74*10³ J

3. Calculate the values of DE and DH for the heating of 55.40 g of xenon from 300 K to 400 K.  Assume ideal-gas behavior and that the heat capacities at constant volume and constant pressure are independent of temperature.

Xenon is a monatomic “ideal” gas, so C_v=3/2*R and C_p=5/2*R (always true for ideal monoatomic gases).  We have the two following equations (always true):
DE=nC_vDT = 526J; DH=nC_pDT = 877 J

4. Calculate the entropy change when 2.0 mol of an ideal gas are allowed to expand isothermally from an initial volume of 1.5 L to 2.4 L.

DS=nR*ln(V₂/V₁) = 7.8 J/K

5. A quantity of 0.50 mol of an ideal gas at 20^oC expands isothermally against a pressure of 2.0 atm from 1.0 L to 5.0 L.  Calculate DS_sys, DS_surr, and DS_univ.

Use the ideal gas law to find initial pressure = 12 atm.  Use the equation:

DS_sys = nR*ln(V₂/V₁) = 6.7 J/K

DS_surr = q_surr/T; DE=0 since DT=0 (it is an isothermal process).  Therefore, w=-q.  We can calculate w=-PDV, and q_sys=-q_surr, so q_surr=w.  We get DS_surr = -2.8 J/K
DS_univ=DS_sys+DS_surr=3.9J/K

6. Supercooled water is liquid water that has been cooled below its normal freezing point.  This state is thermodynamically unstable and tends to freeze into ice spontaneously.  Suppose we have 2.0 mol of supercooled water turning to ice at -10^oC and 1.0 atm.  Calculate DS_sys, DS_surr, and DS_univ for this process (set up a thermodynamic cycle).  C_p water = 75.3 J/K*mol and C_p ice = 37.7 J/K*mol; the molar heat of fusion of water is 6.01 kJ/mol. (DS_fus=n*DH_fus/T, q_surr=n*DH_fus)

Step 1: Water at -10 à Water at 0:

DS = nC_p*ln(T_f/T_i) = 5.6 J/K

Step 2: Water freezes at 0:

DS_fus=DH_fus/T = -44.0 J/K

Step 3: Ice at 0 à Ice at -10:

DS = nC_p*ln(T_f/T_i) = -2.8 J/K

Therefore, DS_tot=DS_sys =5.6 J/K - 44.0 J/K - 2.8 J/K = -41.2 J/K

To find DS_surr, we must find q for each step, add those numbers together and plug into the equation DS_surr = q_rev/T.

Step 1: q_surr=-nC_PDT = -1.5 kJ
Step 2: q_surr=n*DH_fus = 12 kJ
Step 3: q_surr=-nC_pDT = 754 J

q_surr(tot) = 11 kJ, DS_surr=41.8 J/K

DS_univ = DS_sys + DS_surr = 0.6 J/K

7. A 0.590 mol sample of an ideal gas initially at 300 K and 1.50 bar is compressed isothermally to a final pressure of 6.90 bar.  Calculate the change in Gibbs energy for this process.

For an ideal gas, DG = nRT*ln(P₂/P₁) = 2.23 kJ