Chapter 5
Q. 5.1
At room temperature, dry ice (solid CO_{2}) becomes a gas. At 77°F, 13.6 oz of dry ice are put into a steel tank with a volume of 10.00 ft³ . The tank’s pressure gauge registers 11.2 psi. Express the volume (V) of the tank in liters, the amount of CO_{2} in grams and moles (n), the temperature (T) in °C and K and the pressure (P) in bars, mm Hg, and atmospheres.
| ANALYSIS | |
| volume (10.00 ft³); pressure (11.2 psi); temperature (77°F); mass of CO_{2} (13.6 oz) |
Information given: |
| molar mass of CO_{2} Table 1.3: conversion factors for volume and mass formulas for temperature conversion from °F to °C and from °C to K |
Information implied: |
| volume in L pressure in atm, mm Hg, and bar temperature in °C and K moles of CO_{2} |
Asked for: |
Table 1.3 Relations Between Length, Volume, and Mass Units
| Metric | English | Metric-English | |||
| Length | |||||
| 1 km | = 10³ m | 1 ft | = 12 in | 1 in | = 2.54 cm* |
| 1 cm | = 10^{−2} m | 1 yd | = 3 ft | 1 m | = 39.37 in |
| 1 mm | = 10^{−3} m | 1 mi | = 5280 ft | 1 mi | = 1.609 km |
| 1 nm | = 10^{−9} m = 10 Å | ||||
| Volume | |||||
| 1 m³ | = 10^{6}cm^{3} = 10³ L | 1 gal | = 4 qt = 8 pt | 1 ft³ | = 28.32 L |
| 1 cm³ | = 1 mL = 10^{−3} L | 1 qt (U.S. liq) | = 57.75 in³ | 1 L | = 1.057 qt (U.S. liq) |
| Mass | |||||
| 1 kg | = 10³ g | 1 lb | = 16 oz | 1 lb | = 453.6 g |
| 1 mg | = 10^{−3} G | 1 short ton | = 2000 lb | 1 g | = 0.03527 oz |
| 1 metric ton | = 10³ kg | 1 metric ton | = 1.102 short ton | ||
| *This conversion factor is exact; the inch is defined to be exactly 2.54 cm. The other factors listed in this column are approximate, quoted to four significant figures. Additional digits are available if needed for very accurate calculations. For example, the pound is defined to be 453.592 | |||||
STRATEGY
1. Find the necessary conversion factors.
2. Use the temperature conversion formula.
3. Convert oz to grams and use the molar mass of CO_{2} as a conversion factor.
oz → g → mol
Step-by-Step
Verified Solution
volume in L 10.00 ft³ × \frac{28.32 L}{1 ft³} = 283.2 L
pressure in atm 11.2 psi × \frac{1 atm}{14.7 psi} = 0.762 atm
pressure in mm Hg 11.2 psi ×\frac{1 atm}{14.7 psi} × \frac{760 mm Hg}{1 atm} = 579 mm Hg
pressure in bar 11.2 psi ×\frac{1.013 bar}{14.7 psi} = 0.772 bar
temperature in °C °F = 1.8(°C) + 32; 77° = 1.8(°C) + 32°C; °C = 25°C
temperature in K K = (°C) + 273.15; K = 25°C + 273.15 = 298 K
mol CO_{2} 13.6 oz × \frac{1 g}{0.03527 oz} × \frac{1 mol}{44.01 g} = 8.77 mol