What is the new pressure in atm when a constant volume of gas is heated from 25.1 C to 37.5 C the starting pressure is 755.0 mmHg?

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Gay-Lussac’s Law is a direct mathematical relationship. This means there are two connected values and when one (either P or T) goes up, the other (either P or T) also increases. The constant K remains the same value.

The mathematical form of Gay-Lussac’s Law is:

Example #1: 10.0 L of a gas is found to exert 97.0 kPa at 25.0 °C. What would be the required temperature (in Celsius) to change the pressure to standard pressure?

Solution:

1) Change 25.0 °C to 298.0 K and remember that standard pressure in kPa is 101.325. Insert values into the equation to get:

97.0 kPa 101.325 kPa––––––– = –––––––––298.0 K x

x = 311.3 K

2) The question asks for Celsius, so you subtract 273 to get the final answer of 38.3°C. Notice that the volume never enters the problem. This is because the problem is asking about the relationship between pressure and temperature; the volume (as well as the moles) remains constant.

Example #2: What is the new pressure (in atm) when a constant volume of gas is heated from 25.1 °C to 37.5 °C? The starting pressure is 755.0 mmHg.

Solution:

1) Set it up and solve:

755.0 mmHg x–––––––––– = –––––––298.1 K 310.5 K

x = 786.4 mmHg

2) Convert to atm:

(786.4 mmHg) (1 atm / 760.0 mmHg) = 1.035 atm

You could have converted the 755.0 mmHg to atm first, if so desired.

PRACTICE:

If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0 atmospheres and its original temperature was 25.0 °C, what would the final temperature of the gas be? 45.0 °C

Module 8 Science JournalLesson 1 Practice 1 – Interactive Gas Laws PracticeBoyle’s Law Practice:1. His law gives the relationship betweenpressureandvolumeiftemperatureandamountare heldconstant.2. If the volume increases, the pressuredecreases.3. Explain the steps used to solve Example #6:

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