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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