Chapter 7, Problem 7.55P

Interpretation Introduction

(a)

Interpretation:

The following table is to be completed by assuming the gas at a constant temperature.

$V_1$	$T_1$	$V_2$	$T_2$
5.0L	310K	?	250K

Concept Introduction:

According to Charles's law, the volume of gas is directly proportional to the temperature of the gas in K at constant pressure. This relation is represented as

  $V\propto T$

Or,

  $V_2=\frac{V_1{T}_2}{T_1}$

  $T_2=\frac{V_2{T}_1}{V_1}$

Answer to Problem 7.55P

$V_1$	$T_1$	$V_2$	$T_2$
5.0L	310K	4.0L	250K

Explanation of Solution

According to Charles's law, the volume of gas is directly proportional to the Kelvin temperature of the gas at constant pressure. This relation is represented as

  $V\propto T$

  $\frac{V}{T}=constant$

Or,

  $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

  $V_2=\frac{V_1{T}_2}{T_1}$ .......................... (1)

Given that

Initial volume, V₁ = 5.0L

Initial temperature, T₁ = 310K

Final temperature, T₂ = 250K

Put the above values in equ (1)

  $V_2=\frac{5.0L\times 250K}{310K}=4.0L$

The final volume of gas that is V₂ = 4.0L

Thus,

$V_1$	$T_1$	$V_2$	$T_2$
5.0L	310K	4.0L	250K

Interpretation Introduction

(b)

Interpretation:

The following table is to be completed by assuming the gas at a constant temperature.

$V_1$	$T_1$	$V_2$	$T_2$
150mL	45K	?	$45°C$

Concept Introduction:

According to Charles's law, the volume of gas is directly proportional to the temperature of the gas in K at constant pressure. This relation is represented as

  $V\propto T$

Or,

  $V_2=\frac{V_1{T}_2}{T_1}$

  $T_2=\frac{V_2{T}_1}{V_1}$

Answer to Problem 7.55P

$V_1$	$T_1$	$V_2$	$T_2$
150mL	45K	1.1L	$45°C$

Explanation of Solution

According to Charles's law, the volume of gas is directly proportional to the Kelvin temperature of the gas at constant pressure. This relation is represented as

  $V\propto T$

  $\frac{V}{T}=constant$

Or,

  $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

  $V_2=\frac{V_1{T}_2}{T_1}$ .......................... (1)

Given that

Initial volume, V₁ = 150mL

Initial temperature, T₁ = 45K

Final temperature, T₂ = $45°C$

Or,

  $T_2=273+45°C=318K$

Put the above values in equation (1),

  $V_2=\frac{\text{150mL×318K}}{\text{45K}}\text{=1060mL}$

Or,

  $V_2=\text{1060mL×}\frac{\text{1L}}{\text{1000mL}}\text{=1}\text{.06L}$

The final volume of gas that is V₂ = 1.1 L

$V_1$	$T_1$	$V_2$	$T_2$
150mL	45K	1.1L	$45°C$

Interpretation Introduction

(c)

Interpretation:

The following table is to be completed by assuming the gas at a constant temperature.

$V_1$	$T_1$	$V_2$	$T_2$
60.0L	$0.0°C$	180L	?

Concept Introduction:

According to Charles's law, the volume of gas is directly proportional to the temperature of the gas in K at constant pressure. This relation is represented as

  $V\propto T$

Or,

  $V_2=\frac{V_1{T}_2}{T_1}$

  $T_2=\frac{V_2{T}_1}{V_1}$

Answer to Problem 7.55P

$V_1$	$T_1$	$V_2$	$T_2$
60.0L	$0.0°C$	180L	820K

Explanation of Solution

According to Charles's law, the volume of gas is directly proportional to the Kelvin temperature of the gas at constant pressure. This relation is represented as

  $V\propto T$

  $\frac{V}{T}=constant$

Or,

  $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

  $V_2=\frac{V_1{T}_2}{T_1}$ .......................... (1)

Given that −

Initial volume, V₁ = 60L

Final temperature, T₂ = $0.0°C$

Final volume, V₂ = 180L

Or,

  $T_1=273+0.0°C=273K$

  $T_2=\frac{V_2{T}_1}{V_1}$ .......................... (2)

Put the above values in equation (2),

  $T_2=\frac{\text{180L×273K}}{\text{60L}}\text{=819K}$

The final volume of gas that is T₂ = 820K

The final table is −

$V_1$	$T_1$	$V_2$	$T_2$
60.0L	$0.0°C$	180L	820K