# Answers to Gas Laws Practice Problems

1. $$\mathrm{\textrm{molar mass of }Cl_2 = 2(35.45) = 70.90\: g/mole}$$

$$\mathrm{d=\dfrac{mm}{mV}=\dfrac{70.9\:g}{22.4\:L}=3.17\: g/L}$$

1. Molar volume is the volume when $$\mathrm{n = 1.00\: mole}$$.

$$\mathrm{V = ?}$$; $$\mathrm{n = 1.00\: mol}$$; $$\mathrm{T = 78^\circ C + 273 = 351\: K}$$; $$\mathrm{P = 1.20\: atm}$$

$$\mathrm{V=\dfrac{nRT}{P}=\dfrac{(1.00\:mol)(0.08206)(351\:K)}{1.20\:atm}=24.0\: L}$$

1. $$\mathrm{V_1 = 6.66\: L}$$; STP: $$\mathrm{T_1 = 0^\circ C = 273\: K}$$; $$\mathrm{P_1 = 1.00\: atm = 760\: torr}$$;

$$\mathrm{T_2 = 546\: ^\circ C + 273 = 819\: K}$$; $$\mathrm{P_2 = 684\: torr}$$; $$\mathrm{V_2 = ?}$$

$$\mathrm{\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}}$$ $$\mathrm{V_2=V_1\times\dfrac{P_1}{P_2}\times\dfrac{T_2}{T_1}=6.66\:L\times\dfrac{760\:torr}{684\:torr}\times\dfrac{819\:K}{273\:K}=22.2\:L}$$

1. $$\mathrm{mass\: of\: CO_2 = ?}$$; $$\mathrm{V = 5.60\: L}$$; $$\mathrm{T = 273K}$$; $$\mathrm{P = 2.00\: atm}$$

$$\mathrm{n=\dfrac{PV}{RT}=\dfrac{(2.00\:atm)(5.60\:L)}{(0.08206)(273\:K)}=0.500\:moles}$$; $$\mathrm{0.500\: moles\times\dfrac{44.0\:g\:CO_2}{1\:mole\:CO_2} =22.0\: g\: CO_2}$$

1. From the Table, $$\mathrm{P_{H_2O}}$$ at $$\mathrm{21^\circ C = 19\: mm\: Hg}$$; You can subtract in atm or torr.

For torr: $$\mathrm{P_T = 1.02\: atm\times\dfrac{760\:torr}{1\:atm} = 775\: torr}$$; $$\mathrm{P_{N_2} = P_T - P_{H_2O} = 775 - 19 = 756\: torr}$$

For atm: $$\mathrm{P_{H_2O} = 19\: torr\times\dfrac{1\:atm}{760\:torr} = 0.025\: atm}$$; $$\mathrm{P_{N_2} = P_T - P_{H_2O} = 1.02 - 0.025 = 0.99\: atm}$$

1. $$\mathrm{P_{N_2}=0.50\: atm}$$; $$\mathrm{P_{O_2} = 0.30\: atm}$$; $$\mathrm{28.0\: g\: N_2}$$; $$\mathrm{n_T = ?}$$

$$\mathrm{P_T = P_{N_2} + P_{O_2} = 0.50\: atm + 0.30\: atm = 0.80\: atm}$$;

$$\mathrm{moles\: N_2 = 28.0\: g\: N_2 \times\dfrac{1\:mole\:N_2}{28.0\:g\:N_2} = 1.00\: mol\: N_2}$$

$$\mathrm{P_{N_2}=\left(\dfrac{n_{N_2}}{n_T} \right )P_T}$$ $$\mathrm{n_T=\left(\dfrac{P_T}{P_{N_2}}\right)n_{N_2}=\left(\dfrac{0.80}{0.50}\right)1.00\:mol=1.6\:moles}$$

1. $$\mathrm{2.80\: L \times\dfrac{1\:mole\:O_2}{22.4\:L\:O_2}\times\dfrac{4\:mole\:Na}{1\:mole\:O_2}\times\dfrac{23.0\:g\:Na}{1\:mole\:Na}=11.5\: g\: Na}$$
2. First find molar volume at 25°C and 2.00 atm:

$$\mathrm{V=\dfrac{nRT}{P}=\dfrac{(1.00\:mol)(0.08206)(298\:K)}{2.00\:atm}=12.2\:L}$$

$$\mathrm{4.60\: g\: Na \times\dfrac{1\:mole\:Na}{23.0\:g\:Na}\times\dfrac{1\:mole\:O_2}{4\:mole\:Na}\times\dfrac{12.2\:L\:O_2}{1\:mole\:O_2}= 0.610\: L\: O_2}$$

1. First find molar volume at 78°C and 1.20 atm. See problem 2: at 78°C and 1.20 atm, molar volume is 24.0 L/mole.

$$\mathrm{\textrm{molar mass of ethane }(C_2H_6) = 2(12.0) + 6(1.0) = 30.0\: g/mole}$$

$$\mathrm{d=\dfrac{mm}{mV}=\dfrac{30.0\:g}{24.0\:L}= 1.25\: g/L}$$