Chapter U3.10, Problem 7E

(a)

Interpretation Introduction

Interpretation: The relation between P and V as $\text{P = k}\frac{\text{1}}{\text{V}}$ if 1atm = 14.7 lb/in² needs to be explained.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container. The gas pressure is inversely proportional to the volume of gas. This is because as the gas pressure increases, the gas particles come close to each other. It decreases the intermolecular distance between particles and volume decreases.

Answer to Problem 7E

In the given data, the constant value of k proves that:

  $\begin{array}{l}\text{P = k}\frac{\text{1}}{\text{V}}\\ \text{PV}\text{}\text{=k}\end{array}$

Explanation of Solution

Trial	Volume (mL)	Pressure (lb/in2)	Pressure (atm)	$\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{V}$}\right.$	$\begin{array}{l}\text{P = k}\frac{\text{1}}{\text{V}}\\ \text{PV}\text{}\text{=k}\end{array}$
1	60 mL	10 lb/in2	$\frac{{\text{10 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 0}\text{.680 atm}$	$\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{0}\text{.060L}$}\right.\text{= 16}\text{.7}$	$\text{0}\text{.060 L×0}\text{.680 atm = 0}\text{.0408 L}\text{.atm}$
2	40 mL	15 lb/in2	$\frac{{\text{15 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ =1}\text{.02 atm}$	$\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{0}\text{.040L}$}\right.\text{= 25}$	$\text{0}\text{.040 L×1}\text{.02 atm = 0}\text{.0408 L}\text{.atm}$
3	30 mL	20 lb/in2	$\frac{{\text{20 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 1}\text{.36 atm}$	$\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{0}\text{.030L}$}\right.\text{= 33}\text{.3}$	$\text{0}\text{.020 L×1}\text{.36 atm = 0}\text{.0408 L}\text{.atm}$
4	15 mL	40 lb/in2	$\frac{{\text{40 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 2}\text{.72 atm}$	$\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{0}\text{.015 L}$}\right.\text{= 66}\text{.7}$	$\text{0}\text{.015 L×2}\text{.72 atm = 0}\text{.0408 L}\text{.atm}$
5	10 mL	60 lb/in2	$\frac{{\text{60 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 4}\text{.08 atm}$	$\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{0}\text{.010L}$}\right.\text{= 100}$	$\text{0}\text{.010 L×4}\text{.08 atm = 0}\text{.0408 L}\text{.atm}$

The Boyle’s law states that at constant temperature and amount of gas molecules, the volume is inversely proportional to the pressure of gas.

  $\text{Pressure α }\frac{\text{1}}{\text{volume }}$

In the given data, the constant value of k proves that:

  $\begin{array}{l}\text{P = k}\frac{\text{1}}{\text{V}}\\ \text{PV}\text{}\text{=k}\end{array}$

(b)

Interpretation Introduction

Interpretation: The relation between P and V as $\text{PV = k}$ if 1atm = 14.7 lb/in² needs to be explained.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container. The gas pressure is inversely proportional to the volume of gas. This is because as the gas pressure increases, the gas particles come close to each other. It decreases the intermolecular distance between particles and volume decreases.

Answer to Problem 7E

In the given data, the constant value of k proves that:

  $\begin{array}{l}\text{P = k}\frac{\text{1}}{\text{V}}\\ \text{PV}\text{}\text{=k}\end{array}$

Explanation of Solution

The Boyle’s law states that at constant temperature and amount of gas molecules, the volume is inversely proportional to the pressure of gas.

  $\text{Pressure α }\frac{\text{1}}{\text{volume }}$

Trial	Volume (mL)	Pressure (lb/in2)	Pressure (atm)	$\text{PV = k}$
1	60 mL	10 lb/in2	$\frac{{\text{10 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 0}\text{.680 atm}$	$\text{0}\text{.060 L×0}\text{.680 atm = 0}\text{.0408 L}\text{.atm}$
2	40 mL	15 lb/in2	$\frac{{\text{15 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ =1}\text{.02 atm}$	$\text{0}\text{.040 L×1}\text{.02 atm = 0}\text{.0408 L}\text{.atm}$
3	30 mL	20 lb/in2	$\frac{{\text{20 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 1}\text{.36 atm}$	$\text{0}\text{.020 L×1}\text{.36 atm = 0}\text{.0408 L}\text{.atm}$
4	15 mL	40 lb/in2	$\frac{{\text{40 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 2}\text{.72 atm}$	$\text{0}\text{.015 L×2}\text{.72 atm = 0}\text{.0408 L}\text{.atm}$
5	10 mL	60 lb/in2	$\frac{{\text{60 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 4}\text{.08 atm}$	$\text{0}\text{.010 L×4}\text{.08 atm = 0}\text{.0408 L}\text{.atm}$

(c)

Interpretation Introduction

Interpretation: The graph of P versus V needs to be drawn and the relation between gas pressure and volume needs to be explained.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container. The gas pressure is inversely proportional to the volume of gas. This is because as the gas pressure increases, the gas particles come close to each other. It decreases the intermolecular distance between particles and volume decreases.

Answer to Problem 7E

  

The curve between P and V interprets that with increase in pressure of gas, the volume of gas increases.

Explanation of Solution

The Boyle’s law states that at constant temperature and amount of gas molecules, the volume is inversely proportional to the pressure of gas.

  $\text{Pressure α }\frac{\text{1}}{\text{volume }}$

Trial	Volume (mL)	Pressure (lb/in2)	Pressure (atm)	$\text{PV = k}$
1	60 mL	10 lb/in2	$\frac{{\text{10 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 0}\text{.680 atm}$	$\text{0}\text{.060 L×0}\text{.680 atm = 0}\text{.0408 L}\text{.atm}$
2	40 mL	15 lb/in2	$\frac{{\text{15 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ =1}\text{.02 atm}$	$\text{0}\text{.040 L×1}\text{.02 atm = 0}\text{.0408 L}\text{.atm}$
3	30 mL	20 lb/in2	$\frac{{\text{20 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 1}\text{.36 atm}$	$\text{0}\text{.030 L×1}\text{.36 atm = 0}\text{.0408 L}\text{.atm}$
4	15 mL	40 lb/in2	$\frac{{\text{40 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 2}\text{.72 atm}$	$\text{0}\text{.015 L×2}\text{.72 atm = 0}\text{.0408 L}\text{.atm}$
5	10 mL	60 lb/in2	$\frac{{\text{60 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 4}\text{.08 atm}$	$\text{0}\text{.010 L×4}\text{.08 atm = 0}\text{.0408 L}\text{.atm}$

The curve between P and V must be:

  

The curve between P and V interprets that with increase in pressure of gas, the volume of gas increases. Overall pressure is inversely proportional to the volume of gas.

(d)

Interpretation Introduction

Interpretation: The graph of P versus 1/V needs to be drawn and the relation between gas pressure and volume needs to be explained.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container. The gas pressure is inversely proportional to the volume of gas. This is because as the gas pressure increases, the gas particles come close to each other. It decreases the intermolecular distance between particles and volume decreases.

Answer to Problem 7E

  

The curve between P and 1/V interprets that pressure is directly proportional to the inverse volume.

  $\text{Pressure α}\frac{\text{1}}{\text{Volume}}$

Explanation of Solution

Trial	Volume (mL)	Pressure (lb/in2)	Pressure (atm)	$\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{V}$}\right.$
1	60 mL	10 lb/in2	$\frac{{\text{10 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 0}\text{.680 atm}$	$\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{0}\text{.060L}$}\right.\text{= 16}\text{.7}$
2	40 mL	15 lb/in2	$\frac{{\text{15 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ =1}\text{.02 atm}$	$\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{0}\text{.040L}$}\right.\text{= 25}$
3	30 mL	20 lb/in2	$\frac{{\text{20 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 1}\text{.36 atm}$	$\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{0}\text{.030L}$}\right.\text{= 33}\text{.3}$
4	15 mL	40 lb/in2	$\frac{{\text{40 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 2}\text{.72 atm}$	$\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{0}\text{.015 L}$}\right.\text{= 66}\text{.7}$
5	10 mL	60 lb/in2	$\frac{{\text{60 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 4}\text{.08 atm}$	$\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{0}\text{.010L}$}\right.\text{= 100}$

The curve between P and 1/V must be:

  

The curve between P and 1/V interprets that pressure is directly proportional to the inverse volume.

  $\text{Pressure α}\frac{\text{1}}{\text{Volume}}$

(e)

Interpretation Introduction

Interpretation: The volume of gas at 30 lb/in² pressure with the help of given graph needs to be determined.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container. The gas pressure is inversely proportional to the volume of gas. This is because as the gas pressure increases, the gas particles come close to each other. It decreases the intermolecular distance between particles and volume decreases.

Answer to Problem 7E

Volume at 30 lb/in² = 0.020 L = 20.0 mL

Explanation of Solution

Trial	Volume (mL)	Pressure (lb/in2)	Pressure (atm)
1	60 mL	10 lb/in2	$\frac{{\text{10 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 0}\text{.680 atm}$
2	40 mL	15 lb/in2	$\frac{{\text{15 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ =1}\text{.02 atm}$
3	30 mL	20 lb/in2	$\frac{{\text{20 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 1}\text{.36 atm}$
4	15 mL	40 lb/in2	$\frac{{\text{40 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 2}\text{.72 atm}$
5	10 mL	60 lb/in2	$\frac{{\text{60 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 4}\text{.08 atm}$

Convert 30 lb/in² to atm:

1 atm = 14.7 lb/in²

  $\text{1atm ×}\frac{{\text{30 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 2}\text{.04 atm}$

  

Thus the volume of gas at 2.04 atm will be 20.0 mL.

(f)

Interpretation Introduction

Interpretation: The pressure of 50 mL gas with the help of given graph needs to be determined.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container. The gas pressure is inversely proportional to the volume of gas. This is because as the gas pressure increases, the gas particles come close to each other. It decreases the intermolecular distance between particles and volume decreases.

Answer to Problem 7E

The pressure of 0.050 L or 50.0 mL of gas must be 0.82 atm.

Explanation of Solution

Trial	Volume (mL)	Pressure (lb/in2)	Pressure (atm)
1	60 mL	10 lb/in2	$\frac{{\text{10 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 0}\text{.680 atm}$
2	40 mL	15 lb/in2	$\frac{{\text{15 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ =1}\text{.02 atm}$
3	30 mL	20 lb/in2	$\frac{{\text{20 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 1}\text{.36 atm}$
4	15 mL	40 lb/in2	$\frac{{\text{40 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 2}\text{.72 atm}$
5	10 mL	60 lb/in2	$\frac{{\text{60 lb/in}}^{\text{2}}}{\text{14}{\text{.7 lb/in}}^{\text{2}}}\text{ = 4}\text{.08 atm}$

Volume = 50 mL

  $\frac{\text{1}}{\text{Volume}}\text{=}\frac{\text{1}}{\text{0}\text{.050L}}\text{=20}$

  

Thus the pressure of 0.050 L or 50.0 mL of gas must be 0.82 atm.