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Use Gay-Lussac equation to solve the given problem. At 293 K a confined ammonia gas has a pressure of 2.50 atm. At what temperature ( T2 ) would its pressure be equal to 760 mmHg ?

Use Gay-Lussac equation to solve the given problem. At 293 K a confined ammonia gas has a pressure of 2.50 atm. At what temperature ( T2 ) would its pressure be equal to 760 mmHg ?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Use Gay-Lussac equation to solve the given problem.

At 293 K a confined ammonia gas has a pressure of 2.50 atm. At what temperature ( T2 ) would its
pressure be equal to 760 mmHg ?

GAY- LUSSAC'S LAW states at constant volume, the pressure of a fixed mass of gas is directly to the absolute temperature. So, if the temperature increases / decrease, pressure also increases / decreases, P = kTork = P/T. In equation, P,İT; = P2 / T2 or P1 T2 = P2 T1 Sample Problem: 1. The helium tank has a pressure of 650 torr at 25° C. What will be the pressure ( P2 ) if the temperature is tripled? Given : P1- 650 torr T1 = 25° C + 273 = 298 K Formula : P2 = P, T2 + T1 Solution : P2 = ( 650 torr ) ( 348 K) + 298 K Answer : P2 = 759 torr T2 = 3( 25° C ) + 273 = 348 K Unknown : P2 = ? %3D
Transcribed Image Text:GAY- LUSSAC'S LAW states at constant volume, the pressure of a fixed mass of gas is directly to the absolute temperature. So, if the temperature increases / decrease, pressure also increases / decreases, P = kTork = P/T. In equation, P,İT; = P2 / T2 or P1 T2 = P2 T1 Sample Problem: 1. The helium tank has a pressure of 650 torr at 25° C. What will be the pressure ( P2 ) if the temperature is tripled? Given : P1- 650 torr T1 = 25° C + 273 = 298 K Formula : P2 = P, T2 + T1 Solution : P2 = ( 650 torr ) ( 348 K) + 298 K Answer : P2 = 759 torr T2 = 3( 25° C ) + 273 = 348 K Unknown : P2 = ? %3D
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