Chapter 7, Problem 7.56P

Interpretation Introduction

(a)

Interpretation:

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

$V_1$	$T_1$	$V_2$	$T_2$
10.0mL	210K	?	450K

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 7.56P

$V_1$	$T_1$	$V_2$	$T_2$
5.0L	310K	21mL	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 = 10mL

Initial temperature, T₁ = 210K

Final temperature, T₂ = 450K

Put the above values in equation (1),

  $V_2=\frac{\text{10mL×450K}}{\text{210K}}\text{=21mL}$

The final volume of gas that is V₂ = 21mL

The final table is −

	$V_1$	$T_1$	$V_2$	$T_2$
a.	5.0L	310K	21mL	250K

Interpretation Introduction

(b)

Interpretation:

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

$V_1$	$T_1$	$V_2$	$T_2$
255mL	$55°C$	?	150K

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 7.56P

$V_1$	$T_1$	$V_2$	$T_2$
150mL	45K	120mL	$45°C$

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 = 255mL

Initial temperature, T₁ = $55°C$

Final temperature, T₂ = 150K

Or,

  $T_1=273+55°C=328K$

Put the above values in equation (1)

  $V_2=\frac{\text{255mL×150K}}{\text{328K}}\text{=117mL}$

The final volume of gas that is V₂ = 120mL

$V_1$	$T_1$	$V_2$	$T_2$
150mL	45K	120mL	$45°C$

Interpretation Introduction

(c)

Interpretation:

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

$V_1$	$T_1$	$V_2$	$T_2$
13L	$-150°C$	52L	?

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 7.56P

$V_1$	$T_1$	$V_2$	$T_2$
60.0L	$0.0°C$	180L	500K

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 = 13L

Final temperature, T₂ = $-150°C$

Final volume, V₂ = 52L

Or,

  $T_1=273-150°C=123K$

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

Put the above values in equation (2)

  $T_2=\frac{\text{52 L×123 K}}{\text{13 L}}\text{=492 K}$

The final volume of gas that is T₂ = 500K

$V_1$	$T_1$	$V_2$	$T_2$
60.0L	$0.0°C$	180L	500K