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Example: The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft³), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon? Solution: 745 mmHg x 1 atm 760 mmHg n= = 0.980 atm P and I are given in units that are not compatible with the units of the gas constant [R = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres: T=273 +30=303K Substituting these values into the expression we derived for n, we obtain PV RT 0.980 atm x 31150 L atm L 0.08206- mol. K = 1.23 x 10³ mol x 303 K Home Work: Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was -10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in above example.

Example: The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft³), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon? Solution: 745 mmHg x 1 atm 760 mmHg n= = 0.980 atm P and I are given in units that are not compatible with the units of the gas constant [R = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres: T=273 +30=303K Substituting these values into the expression we derived for n, we obtain PV RT 0.980 atm x 31150 L atm L 0.08206- mol. K = 1.23 x 10³ mol x 303 K Home Work: Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was -10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in above example.

Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter5: Gases
Section: Chapter Questions
Problem 52QAP: Nitrogen gas can be obtained by decomposing ammonium nitrate at high temperatures. The nitrogen gas...
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Example: The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft³), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon? Solution: 745 mmHg x 1 atm 760 mmHg P and I are given in units that are not compatible with the units of the gas constant [R = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres: T = 273 +30=303K = = 0.980 atm Substituting these values into the expression we derived for n, we obtain PV RT 0.980 atm x 31150 L atm L mol. K = 1.23 x 10³ mol 0.08206- x 303 K Home Work: Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in above example.
Transcribed Image Text:Example: The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft³), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon? Solution: 745 mmHg x 1 atm 760 mmHg P and I are given in units that are not compatible with the units of the gas constant [R = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres: T = 273 +30=303K = = 0.980 atm Substituting these values into the expression we derived for n, we obtain PV RT 0.980 atm x 31150 L atm L mol. K = 1.23 x 10³ mol 0.08206- x 303 K Home Work: Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in above example.
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