How does the pressure of gas changes if the Kelvin temperature of gas doubles, if the volume and amount of gas are kept constant? cannot be determined It doubles It decreases by half. It increases by half. It quadruples

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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### Understanding Gas Laws: Pressure and Temperature Relationship

#### How does the pressure of gas change if the Kelvin temperature of gas doubles, if the volume and amount of gas are kept constant?

- ○ Cannot be determined
- ○ It doubles
- ○ It decreases by half
- ○ It increases by half
- ○ It quadruples

#### Explanation

According to Gay-Lussac's Law, when the volume and the amount of gas are kept constant, the pressure of a gas is directly proportional to its temperature in Kelvin. Mathematically, it can be expressed as:

\[ \frac{P1}{T1} = \frac{P2}{T2} \]

Where:
- \( P1 \) and \( P2 \) represent the initial and final pressures, respectively.
- \( T1 \) and \( T2 \) represent the initial and final temperatures in Kelvin, respectively.

From this relationship, if the temperature \( T2 \) is twice the initial temperature \( T1 \) (i.e., \( T2 = 2T1 \)), then the pressure will also double. Therefore, the correct answer is:

- ○ It doubles
Transcribed Image Text:### Understanding Gas Laws: Pressure and Temperature Relationship #### How does the pressure of gas change if the Kelvin temperature of gas doubles, if the volume and amount of gas are kept constant? - ○ Cannot be determined - ○ It doubles - ○ It decreases by half - ○ It increases by half - ○ It quadruples #### Explanation According to Gay-Lussac's Law, when the volume and the amount of gas are kept constant, the pressure of a gas is directly proportional to its temperature in Kelvin. Mathematically, it can be expressed as: \[ \frac{P1}{T1} = \frac{P2}{T2} \] Where: - \( P1 \) and \( P2 \) represent the initial and final pressures, respectively. - \( T1 \) and \( T2 \) represent the initial and final temperatures in Kelvin, respectively. From this relationship, if the temperature \( T2 \) is twice the initial temperature \( T1 \) (i.e., \( T2 = 2T1 \)), then the pressure will also double. Therefore, the correct answer is: - ○ It doubles
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