“The final volume of the gas” should be calculated Concept introduction: Here we are using the concept of different gaseous law which are combined together to form a combined gaseous law. ► Boyle’s Law: It states that at constant temperature, the pressure of the gas has inverse relation with volume P α 1 V ► Charles’ Law: It states that at constant pressure, the volume is directly proportional to the temperature V α T ► Gay-Lussac’s law: It states that at constant volume, the pressure of the gas has direct relation with temperature P α T ► Now by combining the three law we get a relationship between pressure, volume and temperature which is known as combined gas equation P 1 V 1 T 1 = P 2 V 2 T 2 Given: Volume of the gas, V 1 = 780 mL Pressure, P 1 =1 atm Temperature, T 1 = 37 0 C Pressure, P 2 = 675 mmHg Temperature, T 2 = 22 0 C

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Answer 1

Question

Chapter 8, Problem 8.96CQ

Interpretation Introduction

Interpretation:

“The final volume of the gas” should be calculated

Concept introduction:

Here we are using the concept of different gaseous law which are combined together to form a combined gaseous law.

► Boyle’s Law: It states that at constant temperature, the pressure of the gas has inverse relation with volume

$P α \frac{1}{V}$

► Charles’ Law: It states that at constant pressure, the volume is directly proportional to the temperature

$V α T$

► Gay-Lussac’s law: It states that at constant volume, the pressure of the gas has direct relation with temperature

$P α T$

► Now by combining the three law we get a relationship between pressure, volume and temperature which is known as combined gas equation

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

Given:

Volume of the gas, $V_{1} = 780 mL$

Pressure, $P_{1} =1 atm$

Temperature, $T_{1} = 37^{0} C$

Pressure, $P_{2} = 675 mmHg$

Temperature, $T_{2} {= 22}^{0} C$

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Answer 2

Question

Chapter 8, Problem 8.96CQ

Interpretation Introduction

Interpretation:

“The final volume of the gas” should be calculated

Concept introduction:

Here we are using the concept of different gaseous law which are combined together to form a combined gaseous law.

► Boyle’s Law: It states that at constant temperature, the pressure of the gas has inverse relation with volume

$P α \frac{1}{V}$

► Charles’ Law: It states that at constant pressure, the volume is directly proportional to the temperature

$V α T$

► Gay-Lussac’s law: It states that at constant volume, the pressure of the gas has direct relation with temperature

$P α T$

► Now by combining the three law we get a relationship between pressure, volume and temperature which is known as combined gas equation

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

Given:

Volume of the gas, $V_{1} = 780 mL$

Pressure, $P_{1} =1 atm$

Temperature, $T_{1} = 37^{0} C$

Pressure, $P_{2} = 675 mmHg$

Temperature, $T_{2} {= 22}^{0} C$

Expert Solution & Answer

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Answer 3

Question

Chapter 8, Problem 8.96CQ

Interpretation Introduction

Interpretation:

“The final volume of the gas” should be calculated

Concept introduction:

Here we are using the concept of different gaseous law which are combined together to form a combined gaseous law.

► Boyle’s Law: It states that at constant temperature, the pressure of the gas has inverse relation with volume

$P α \frac{1}{V}$

► Charles’ Law: It states that at constant pressure, the volume is directly proportional to the temperature

$V α T$

► Gay-Lussac’s law: It states that at constant volume, the pressure of the gas has direct relation with temperature

$P α T$

► Now by combining the three law we get a relationship between pressure, volume and temperature which is known as combined gas equation

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

Given:

Volume of the gas, $V_{1} = 780 mL$

Pressure, $P_{1} =1 atm$

Temperature, $T_{1} = 37^{0} C$

Pressure, $P_{2} = 675 mmHg$

Temperature, $T_{2} {= 22}^{0} C$

Expert Solution & Answer

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Check out a sample textbook solution

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Answer 4

Question

Chapter 8, Problem 8.96CQ

Interpretation Introduction

Interpretation:

“The final volume of the gas” should be calculated

Concept introduction:

Here we are using the concept of different gaseous law which are combined together to form a combined gaseous law.

► Boyle’s Law: It states that at constant temperature, the pressure of the gas has inverse relation with volume

$P α \frac{1}{V}$

► Charles’ Law: It states that at constant pressure, the volume is directly proportional to the temperature

$V α T$

► Gay-Lussac’s law: It states that at constant volume, the pressure of the gas has direct relation with temperature

$P α T$

► Now by combining the three law we get a relationship between pressure, volume and temperature which is known as combined gas equation

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

Given:

Volume of the gas, $V_{1} = 780 mL$

Pressure, $P_{1} =1 atm$

Temperature, $T_{1} = 37^{0} C$

Pressure, $P_{2} = 675 mmHg$

Temperature, $T_{2} {= 22}^{0} C$

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Chapter 8, Problem 8.96CQ

Interpretation Introduction

Interpretation:

“The final volume of the gas” should be calculated

Concept introduction:

Here we are using the concept of different gaseous law which are combined together to form a combined gaseous law.

► Boyle’s Law: It states that at constant temperature, the pressure of the gas has inverse relation with volume

$P α \frac{1}{V}$

► Charles’ Law: It states that at constant pressure, the volume is directly proportional to the temperature

$V α T$

► Gay-Lussac’s law: It states that at constant volume, the pressure of the gas has direct relation with temperature

$P α T$

► Now by combining the three law we get a relationship between pressure, volume and temperature which is known as combined gas equation

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

Given:

Volume of the gas, $V_{1} = 780 mL$

Pressure, $P_{1} =1 atm$

Temperature, $T_{1} = 37^{0} C$

Pressure, $P_{2} = 675 mmHg$

Temperature, $T_{2} {= 22}^{0} C$

Answer 6

Chapter 8, Problem 8.96CQ

Interpretation Introduction

Interpretation:

“The final volume of the gas” should be calculated

Concept introduction:

Here we are using the concept of different gaseous law which are combined together to form a combined gaseous law.

► Boyle’s Law: It states that at constant temperature, the pressure of the gas has inverse relation with volume

$P α \frac{1}{V}$

► Charles’ Law: It states that at constant pressure, the volume is directly proportional to the temperature

$V α T$

► Gay-Lussac’s law: It states that at constant volume, the pressure of the gas has direct relation with temperature

$P α T$

► Now by combining the three law we get a relationship between pressure, volume and temperature which is known as combined gas equation

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

Given:

Volume of the gas, $V_{1} = 780 mL$

Pressure, $P_{1} =1 atm$

Temperature, $T_{1} = 37^{0} C$

Pressure, $P_{2} = 675 mmHg$

Temperature, $T_{2} {= 22}^{0} C$

Answer 7

Chapter 8, Problem 8.96CQ

Interpretation Introduction

Interpretation:

“The final volume of the gas” should be calculated

Concept introduction:

Here we are using the concept of different gaseous law which are combined together to form a combined gaseous law.

► Boyle’s Law: It states that at constant temperature, the pressure of the gas has inverse relation with volume

$P α \frac{1}{V}$

► Charles’ Law: It states that at constant pressure, the volume is directly proportional to the temperature

$V α T$

► Gay-Lussac’s law: It states that at constant volume, the pressure of the gas has direct relation with temperature

$P α T$

► Now by combining the three law we get a relationship between pressure, volume and temperature which is known as combined gas equation

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

Given:

Volume of the gas, $V_{1} = 780 mL$

Pressure, $P_{1} =1 atm$

Temperature, $T_{1} = 37^{0} C$

Pressure, $P_{2} = 675 mmHg$

Temperature, $T_{2} {= 22}^{0} C$

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 8.1 - Prob. 8.1QAP Ch. 8.1 - Prob. 8.2QAP Ch. 8.1 - Prob. 8.3QAP Ch. 8.1 - Prob. 8.4QAP Ch. 8.1 - Prob. 8.5QAP Ch. 8.1 - Prob. 8.6QAP Ch. 8.1 - Prob. 8.7QAP Ch. 8.1 - Prob. 8.8QAP Ch. 8.2 - Why do scuba divers need to exhale air when they...Ch. 8.2 - Why does a sealed bag of chips expand when you...

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