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Answers to More Chapter 05 Study Questions

  1. \(\mathrm{V_1 = 24.0\: L}\); \(\mathrm{T_1 = 20^\circ C = 293\: K}\); \(\mathrm{P_1 = 1.50\: atm}\).  \(\mathrm{V_2 = 36.0\: L}\); \(\mathrm{T_2 = 313^\circ C = 586\: K}\); \(\mathrm{P_2 = ?}\)

\(\mathrm{\dfrac{P_1\times V_1}{T_1}=\dfrac{P_2\times V_2}{T_2}}\);  \(\mathrm{P_2=P_1\times\dfrac{V_1}{V_2}\times\dfrac{T_2}{T_1}=1.50\:atm\times\dfrac{24\:L}{36\:L}\times\dfrac{586\:K}{293\:K}=2.00\: atm}\)

 

  1. \(\mathrm{P_{N_2} = 0.50\: atm}\); \(\mathrm{P_{O_2} = 0.20\: atm}\); \(\mathrm{P_T = 0.80\: atm}\)
  1. \(\mathrm{P_{CO_2} = ?}\);  \(\mathrm{P_{CO_2} = P_T - (P_{N_2} + P_{O_2}) = 0.80 - (0.50 + 0.20) = 0.10\: atm}\)
  2. \(\mathrm{n_{N_2} = 0.25\: moles}\); \(\mathrm{n_{O_2} = ?}\)      \(\mathrm{\dfrac{n_{O_2}}{n_{N_2}}=\dfrac{P_{O_2}}{P_{N_2}}}\); \(\mathrm{n_{O_2}=\dfrac{P_{O_2}}{P_{N_2}}\times n_{N_2}=\dfrac{0.20\:atm}{0.50\:atm}\times0.25\:mol=0.10\:mol}\)

 

  1. \(\mathrm{pentane = C_5H_{12}}\); \(\mathrm{mass = ?}\), \(\mathrm{V = 11.2\: L}\), \(\mathrm{T = 273\: K}\), \(\mathrm{P = 2.40\: atm}\); so find n first.

\(\mathrm{n=\dfrac{PV}{RT}=\dfrac{2.40\:atm\times11.2\:L}{0.08206\times273\:K}=1.20\:mol}\); \(\mathrm{1.20\:mol\times\dfrac{72.0\:g}{1\:mol}=86.4\: g}\)

 

    1. \(\mathrm{N_2(g) + 3 H_2(g)\rightarrow 2 NH_3(g)}\)
    2. \(\mathrm{4.50\: L\: H_2\times\dfrac{2\:L\:NH_3}{3\:L\:H_2}=3.00\: L\: NH_3}\)
    3. \(\mathrm{5.60\: L\: N_2\times\dfrac{1\:mol\:N_2}{22.4\:L\:N_2}\times\dfrac{2\:mol\:NH_3}{1\:mol\:N_2}\times\dfrac{17.0\:g\:NH_3}{1\:mol\:NH_3}=8.50\: g\: NH_3}\)
    4. first find molar volume @ 25°C, 1 atm:  \(\mathrm{V=\dfrac{nRT}{P}=\dfrac{(1\:mol)(0.08206)(298K)}{1\:atm}= 24.5\: L}\)

\(\mathrm{12.1\: g\: H_2\times \dfrac{1\:mol\:H_2}{2.016\:g\:H_2} \times \dfrac{2\:mol\:NH_3}{3\:mol\:H_2}\times\dfrac{24.5\:L\:NH_3}{1\:mol\:NH_3}=98.0\: L}\)

 

  1. \(\mathrm{d=\dfrac{mm}{mV}}\); \(\mathrm{T=65^\circ C=338K}\); \(\mathrm{P=745\:mmHg\times\dfrac{1\:atm}{760\:mmHg}=0.980\: atm}\); \(\mathrm{mm=44.0\:g}\)

\(\mathrm{mV=\dfrac{nRT}{P}=\dfrac{(1\:mol)(0.08206)(338K)}{0.980\:atm}=28.3\:L}\);  \(\mathrm{d=\dfrac{44.0\:g}{28.3\:L}=1.55\: g/L}\)

     

  1. \(\mathrm{d=\dfrac{mm}{mV}}\); \(\mathrm{mm=d\times mV}\); \(\mathrm{T=34^\circ C=307K}\)

\(\mathrm{mV=\dfrac{nRT}{P}=\dfrac{(1\:mol)(0.08206)(307K)}{1.26\:atm}=20.0\:L}\); \(\mathrm{mm=\dfrac{1.50\:g}{L} \times 20.0\:L=30.0\: g}\)

 

  1. \(\mathrm{V_1 = 100\: L}\); \(\mathrm{T_1 = 12^\circ C = 293\: K}\); \(\mathrm{P_1(dry\: gas) = 758\: mmHg}\).  \(\mathrm{V_2 = ?}\); \(\mathrm{T_2 = 20^\circ C = 293\: K}\); \(\mathrm{P_2(wet\: gas) = 740.\: mmHg}\).  

From Table:  \(\mathrm{P_{H_2O}(20^\circ C) = 17.5\: mmHg}\)

\(\mathrm{P_2(wet\: gas) = P_2(dry\: gas) + P_{H_2O}}\);  \(\mathrm{P_2(dry\: gas) = 740.\: mmHg - 17.5\: mmHg = 723\: mmHg}\)

\(\mathrm{\dfrac{P_1\times V_1}{T_1}=\dfrac{P_2\times V_2}{T_2}}\);   \(\mathrm{V_2=V_1\times\dfrac{P_1}{P_2}\times\dfrac{T_2}{T_1}=100\:L\times\dfrac{758}{723}\times\dfrac{293\:K}{285\:K}=108\: L}\)

 

  1. Calculate the final partial pressure of each gas and add them.

\(\ce{O2}\):  \(\mathrm{V_1 = 0.20\: L}\); \(\mathrm{T_1 = 0^\circ C = 273\: K}\);  \(\mathrm{P_1 = 1.0\: atm}\); \(\mathrm{V_2 = 0.40\: L}\); \(\mathrm{T_2 = T_1}\)  \(\mathrm{P_2 = ?}\)

\(\ce{N2}\):  \(\mathrm{V_1 = 0.10\: L}\); \(\mathrm{T_1 = 0^\circ C = 273\: K}\);  \(\mathrm{P_1 = 2.0\: atm}\); \(\mathrm{V_2 = 0.40\: L}\); \(\mathrm{T_2 = T_1}\)  \(\mathrm{P_2 = ?}\)

\(\mathrm{P_1 \times V_1 = P_2 \times V_2}\);  \(\mathrm{P_2=P_1 \times \dfrac{V_1}{V_2}}\)

for \(\ce{O2}\):  \(\mathrm{P_2=1.0\:atm\times\dfrac{0.20\:L}{0.40\:L}=0.50\: atm}\);   for \(\ce{N2}\):  \(\mathrm{P_2=2.0\:atm\times\dfrac{0.10\:L}{0.40\:L}=0.50\: atm}\)

\(\mathrm{P_T = P_{O_2} + P_{N_2} = 0.50\: atm + 0.50\: atm = 1.00\: atm}\)