Chapter 7, Problem 44P

Interpretation Introduction

(a)

Interpretation:

The following table is to be completed by assuming a fixed amount of gas:

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
0.55atm	1.1L	340K	?	3.0L	298K

Concept Introduction:

The concept of the combination gas will be used

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Or,

  $P_2=\frac{P_1{V}_1{T}_2}{T_1{V}_2}$

Or,

  $V_2=\frac{P_1{V}_1{T}_2}{T_1{P}_2}$

Or,

  $T_2=\frac{V_2{T}_1{P}_2}{P_1{V}_1}$

Here,

  $P_1$ = initial pressure of the gas

  $V_1$ = initial volume of the gas

  $T_1$ = initial temperature of the gas

  $P_2$ = final pressure of the gas

  $V_2$ =final volume of the gas

  $T_2$ = final temperature of the gas

Answer to Problem 44P

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
0.55atm	1.1L	340K	0.18atm	3.0L	298K

Explanation of Solution

The combined gas law is used to determine the unknown values. The combined gas law includes the change in pressure-temperature and the volume of the gas. This relation is represented as

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$ .......................... (1)

Given that −

Initial volume, V₁ = 1.1L

Initial temperature, T₁ = 340K

Initial pressure, P₁ = 0.55atm

Final temperature, T₂ = 298K

Final volume, V₂ = 3.0L

  $P_2=\frac{P_1{V}_1{T}_2}{T_1{V}_2}$

Put the values in above equation.

  $P_2=\frac{0.55atm\times 1.1L\times 298K}{340K\times 3.0L}=0.18atm$

The final pressure of gas that is P₂ = 0.18atm.

Thus,

	$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
a.	0.55atm	1.1L	340K	0.18atm	3.0L	298K

Interpretation Introduction

(b)

Interpretation:

The following table is to be completed by assuming a fixed amount of gas:

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
735mmHg	1.2L	298K	1.1atm	?	$0.0°C$

Concept Introduction:

The concept of the combination gas will be used

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Or,

  $P_2=\frac{P_1{V}_1{T}_2}{T_1{V}_2}$

Or,

  $V_2=\frac{P_1{V}_1{T}_2}{T_1{P}_2}$

Or,

  $T_2=\frac{V_2{T}_1{P}_2}{P_1{V}_1}$

Here,

  $P_1$ = initial pressure of the gas

  $V_1$ = initial volume of the gas

  $T_1$ = initial temperature of the gas

  $P_2$ = final pressure of the gas

  $V_2$ =final volume of the gas

  $T_2$ = final temperature of the gas

Answer to Problem 44P

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
735mmHg	1.2L	298K	1.1atm	0.89L	$0.0°C$

Explanation of Solution

The combined gas law is used to determine the unknown values. The combined gas law includes the change in pressure-temperature and the volume of the gas. This relation is represented as

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$ .......................... (1)

Given that

Initial volume, V₁ = 1.2L

Initial temperature, T₁ = 298

Initial pressure, P₁ = 735mmHg

Final pressure, P₂ = 1.1atm

Final temperature, T₂ = $0.0°C$

Or,

  $T_2=273+0.0°C=273K$

  $V_2=\frac{P_1{V}_1{T}_2}{T_1{P}_2}$

Put the values in the above equation.

  $V_2=\frac{\text{1}\text{.1L×}\left(\text{735mmHg×}\frac{\text{1atm}}{\text{760mmHg}}\right)\text{×273K}}{\text{298K×1}\text{.1atm}}\text{=0}\text{.89L}$

The final volume of gas that is V₂ = 0.89L.

Thus,

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
735mmHg	1.2L	298K	1.1atm	0.89L	$0.0°C$

Interpretation Introduction

(c)

Interpretation:

The following table is to be completed by assuming a fixed amount of gas.

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
7.5atm	230mL	$-120°C$	15atm	0.45L	?

Concept Introduction:

The concept of the combination gas will be used

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Or,

  $P_2=\frac{P_1{V}_1{T}_2}{T_1{V}_2}$

Or,

  $V_2=\frac{P_1{V}_1{T}_2}{T_1{P}_2}$

Or,

  $T_2=\frac{V_2{T}_1{P}_2}{P_1{V}_1}$

Here,

  $P_1$ = initial pressure of the gas

  $V_1$ = initial volume of the gas

  $T_1$ = initial temperature of the gas

  $P_2$ = final pressure of the gas

  $V_2$ =final volume of the gas

  $T_2$ = final temperature of the gas

Answer to Problem 44P

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
7.5atm	230mL	$-120°C$	15atm	0.45L	600K

Explanation of Solution

The combined gas law is used to determine the unknown values. The combined gas law includes the change in pressure-temperature and the volume of the gas. This relation is represented as −

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$ .......................... (1)

Given that −

Initial volume, V₁ = 230mL

Initial pressure, P₁ = 7.5atm

Initial temperature, T₁ = $-120.0°C$

  $T_1=273-120.0°C=153K$

Final pressure, P₂ = 15atm

Final volume, V₂ = 0.45L

  $T_2=\frac{V_2{T}_1{P}_2}{P_1{V}_1}$

Put the values in the above equation.

  $T_2=\frac{0.45L\times 153K\times 15atm}{7.5atm\times 0.230L}=6.0\times {10}^{2}K$

The final volume of gas that is T₂ = 600K.

Thus,

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
7.5atm	230mL	$-120°C$	15atm	0.45L	600K