Chapter 7, Problem 7.59P

Interpretation Introduction

(a)

Interpretation:

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

$P_1$	$T_1$	$P_2$	$T_2$
3.25atm	298K	?	398K

Concept Introduction:

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

  $P\propto T$

Or,

  $P_2=\frac{P_1{T}_2}{T_1}$

  $T_2=\frac{P_2{T}_1}{P_1}$

Answer to Problem 7.59P

$P_1$	$T_1$	$P_2$	$T_2$
5.0L	310K	4.34atm	250K

Explanation of Solution

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

  $P\propto T$

  $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

  $P_2=\frac{P_1{T}_2}{T_1}$ .......................... (1)

For (a) −

Given that −

Initial pressure, P₁ = 3.25atm

Initial temperature, T₁ = 298K

Final temperature, T₂ = 398K

Put the above values in equation (1)

  $P_2=\frac{\text{5}\text{.0L×398K}}{\text{298K}}\text{=4}\text{.34atm}$

The final pressure of gas that is P₂ = 4.34atm

Thus,

$P_1$	$T_1$	$P_2$	$T_2$
5.0L	310K	4.34atm	250K

Interpretation Introduction

(b)

Interpretation:

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

$P_1$	$T_1$	$P_2$	$T_2$
550mmHg	273K	?	$-100°C$

Concept Introduction:

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

  $P\propto T$

Or,

  $P_2=\frac{P_1{T}_2}{T_1}$

  $T_2=\frac{P_2{T}_1}{P_1}$

Answer to Problem 7.59P

$P_1$	$T_1$	$P_2$	$T_2$
150mL	45K	349mmHg	$45°C$

Explanation of Solution

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

  $P\propto T$

  $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

  $P_2=\frac{P_1{T}_2}{T_1}$ .......................... (1)

Given that −

Initial pressure, P₁ = 550mmHg

Initial temperature, T₁ = 273K

Final temperature, T₂ = $-100°C$

Or,

  $T_2=273-100°C=173K$

Put the above values in equation (1)

  $P_2=\frac{\text{550mL×173K}}{\text{273K}}\text{=349mmHg}$

The final pressure of gas that is P₂ = 349mmHg

$P_1$	$T_1$	$P_2$	$T_2$
150mL	45K	349mmHg	$45°C$

Interpretation Introduction

(c)

Interpretation:

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

$P_1$	$T_1$	$P_2$	$T_2$
0.25atm	$250°C$	955mmHg	?

Concept Introduction:

The gas at constant volume. This relation is represented as

  $P\propto T$

Or,

  $P_2=\frac{P_1{T}_2}{T_1}$

  $T_2=\frac{P_2{T}_1}{P_1}$

Answer to Problem 7.59P

$P_1$	$T_1$	$P_2$	$T_2$
60.0L	$0.0°C$	180L	1300K

Explanation of Solution

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

  $P\propto T$

  $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

  $P_2=\frac{P_1{T}_2}{T_1}$ .......................... (1)

Given that −

Initial volume, P₁ = 0.50atm

Or,

1 atm = 760mm Hg

  ${\text{Hence,P}}_{\text{1}}\text{=0}\text{.50atm×}\frac{\text{760mmHg}}{\text{1atm}}\text{=380mmHg}$

Final temperature, T₁ = $250.0°C$

  $T_1=273+250°C=523K$

Final volume, P₂ = 955mmHg

  $T_2=\frac{P_2{T}_1}{P_1}$ .......................... (2)

Put the above values in equation (2)

  $T_2=\frac{523K\times 955mmHg}{380mmHg}=1300K$

The final temperature of the gas that is T₂ = 1300K

$P_1$	$T_1$	$P_2$	$T_2$
60.0L	$0.0°C$	180L	1300K