# Answers to More Chapter 05 Study Questions

1.  V1 = 24.0 L; T1 = 20°C = 293 K; P1 = 1.50 atm.  V2 = 36.0 L; T2 = 313°C = 586 K; P2 = ?

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

2.  $$\mathrm{P_{N_2} = 0.50\: atm}$$; $$\mathrm{P_{O_2} = 0.20\: atm}$$; PT = 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. $$\mathrm{n_{N_2} = 0.25\: moles}$$; $$\mathrm{n_{O_2} = ?}$$      $$\mathrm{\dfrac{n_{O_2}}{n_{N_2}}=\dfrac{P_{O_2}}{P_{N_2}}}$$; $$\mathrm{n_{O_2}=\dfrac{P_{O_2}}{P_{N_2}}\times n_{N_2}=\dfrac{0.20\:atm}{0.50\:atm}\times0.25\:mol=0.10\:mol}$$

3.  pentane = C5H12; mass = ?, V = 11.2 L, T = 273 K, P = 2.40 atm; so find n first.

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

4.

1. N2(g)  +  3 H2(g)       2 NH3(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:  $$\mathrm{V=\dfrac{nRT}{P}=\dfrac{(1\:mol)(0.08206)(298K)}{1\:atm}= 24.5\: L}$$

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

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

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

6.  $$\mathrm{d=\dfrac{mm}{mV}}$$; $$\mathrm{mm=d\times mV}$$; $$\mathrm{T=34^\circ C=307K}$$

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

7.  V1 = 100 L; T1 = 12°C = 293 K; P1(dry gas) = 758 mmHg.  V2 = ?; T2 = 20°C = 293 K; P2(wet gas) = 740. mmHg.

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

$$\mathrm{P_2(wet\: gas) = P_2(dry\: gas) + P_{H_2O}}$$;  P2(dry gas) = 740. mmHg – 17.5 mmHg = 723 mmHg

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

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

O2:  V1 = 0.20 L; T1 = 0°C = 273 K;  P1 = 1.0 atm; V2 = 0.40 L; T2 = T1  P2 = ?

N2:  V1 = 0.10 L; T1 = 0°C = 273 K;  P1 = 2.0 atm; V2 = 0.40 L; T2 = T1  P2 = ?

P1 x V1 = P2 x V2;  $$\mathrm{P_2=P_1 \times \dfrac{V_1}{V_2}}$$

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

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