Chapter 7, Problem 43P

Interpretation Introduction

(a)

Interpretation:

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

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
0.90atm	4.0L	265K	?	3.0L	310K

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 43P

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
0.90atm	4.0L	265K	1.4atm	3.0L	310K

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 = 4.0L

Initial temperature, T₁ = 265K

Initial pressure, P₁ = 0.90atm

Final temperature, T₂ = 310K

Final volume, V₂ = 3.0L

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

Put the values in the above equation.

  $P_2=\frac{\text{0}\text{.90atm×4}\text{.0L×310K}}{\text{265K×3}\text{.0L}}\text{=1}\text{.4atm}$

The final pressure of gas that is P₂ = 1.4atm

Thus,

	$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
a.	0.90atm	4.0L	265K	1.4atm	3.0L	310K

Interpretation Introduction

(b)

Interpretation:

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

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
1.2atm	75L	$5.0°C$	700mmHg	?	$50°C$

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 43P

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
1.2atm	75L	$5.0°C$	700mmHg	110L	$50°C$

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 = 75L

Initial temperature, T₁ = $5.0°C$

  $T_1=273+5.0°C=278K$

Initial pressure, P₁ = 1.2atm

Final pressure, P₂ = 700mmHg

Final temperature, T₂ = $50°C$

Or,

  $T_2=273+50°C=323K$

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

Put the values in the above equation.

  $V_2=\frac{\text{75L×1}\text{.2atm×323K}}{\text{278K×}\left(\text{700mmHg×}\frac{\text{1atm}}{\text{760mmHg}}\right)}\text{=1}{\text{.1×10}}^{\text{2}}\text{L}$

The final volume of gas that is V₂ = $1.1\times {10}^{2}L$

Thus,

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
1.2atm	75L	$5.0°C$	700mmHg	110L	$50°C$

Interpretation Introduction

(c)

Interpretation:

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

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
200mmHg	125mL	298K	100mmHg	0.62L	?

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 43P

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
200mmHg	125mL	298K	100mmHg	0.62L	740K

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 = 125mL

Initial pressure, P₁ = 200mmHg

Initial temperature, T₁ = 298K

Final pressure, P₂ = 100mmHg

Final volume, V₂ = 0.62L

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

Put the values in the above equation.

  $T_2=\frac{\text{0}\text{.62L×298K×100mmHg}}{\text{200mmHg×0}\text{.125L}}\text{=7}{\text{.4×10}}^{\text{2}}\text{K}$

The final volume of gas that is T₂ = 740K

The final table is

$P_1$	$V_1$	$T_1$	$P_2$	$V_2$	$T_2$
200mmHg	125mL	298K	100mmHg	0.62L	740K