Homework Solutions

February 2

Chapter 5

26.    a.    horizontal line (PV is constant if T is constant)   
        b.    line with positive slope
        c.    line with positive slope
        d.    curve with continuously decreasing positive slope
        e.    line with positive slope
        f.    horizontal line

33.    P(1)/T(1) = P(2)/T(2);  P(1) = 40.0 atm;  T(1) = 0 C = 273 K

a.    T(2) = 125 C = 398 K; P(2) = P(1)*T(2)/T(1) = 58.3 atm
b.    P(2) = 150. atm; T(2) = P(2)*T(1)/P(1) = 1020 K
c.    P(2) = 25.0 atm; T(2) = 171 K

31.    P(1)*V(1)/T(1) = P(2)*V(2)/T(2);  P(2) = P(1)*V(1)*T(2)/(T(1)*V(2))

P(1) = 710 Torr; V(1) = 500 mL; T(1) = 30. C = 303 K; T(2) = 1093 K; V(2) = 25. mL

P(2) = 51000 Torr!!!

27.    Original pressure of H₂ = 475 Torr in 2.00 L
Final volume of H₂ = 3.00 L
P(i)*V(i) = P(f)*V(f); P(f) = P(i)*V(i)/V(f) = 317 Torr

        Original pressure of N₂ = 0.200 atm in 1.00 L
Final volume of N₂ = 3.00 L
P(f) = 0.067 atm = 50.7 Torr

Total pressure = 317 torr + 50.7 torr = 368 torr

43.    P_total = P_CH4 + P_O2 = 0.175 atm + 0.250 atm = 0.425 atm
 

a.    x_CH4 = n_CH4/(n_CH4 + n_O2) = P_CH4/P_total = 0.412
      x_O2 = n_O2/(n_CH4 + n_O2) = P_O2/P_total = 0.588

b.    n_tot = (0.425 atm)*(2.00 L)/((0.08206 L atm/mol K)*(273 + 65)) = 0.0306 moles

c.    n_O2 = x_O2 * P_total = 0.588 * 0.0306 moles = 0.01799 moles O₂

                18.016 g/mol * 0.01799 moles O₂ = 0.324 g O₂

     n_CH4 = x_CH4 * P_total = 0.412 * 0.0306 moles = 0.0126 moles CH₄

                16.043 g/mol * 0.0126 moles CH₄ = 0.202 g CH₄

41.    P(total) = P(He) + P(water) = 1.00 atm
P(water) = 23.8 Torr = 0.0313 atm; 
P(He) = P(total) - P(water) = 0.969 atm
n(He) = 0.586 g He/4.003 g/mole = 0.146 mole He
V = n(He)*R*T/P(He) = 3.68 L