Chapter 5.3, Problem 1RC

(a)

Interpretation Introduction

Interpretation:

The change in volume when the temperature of a gas is doubled at constant pressure has to be calculated.

Concept Introduction:

STP in chemistry is the short form for Standard Temperature and Pressure. STP most usually is used when performing calculations on gases, such as gas density.

Use the combined gas law

$\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}$

$\begin{array}{l}\text{where,}{\text{P}}_{\text{1}}\text{,}{\text{V}}_{\text{1}}\text{ and}{\text{T}}_{\text{1}}\text{are}\text{the}\text{initial}\text{}\text{pressure,}\text{volume,and}\text{temperature}\text{.}\\ {\text{P}}_{\text{2}}{\text{,V}}_{\text{2}}\text{and}{\text{T}}_{\text{2}}\text{are}\text{the}\text{final}\text{pressure,}\text{volume,}\text{and}\text{temperature}\text{.}\\ \text{pressure}\text{remains}\text{same}\text{. so,}\\ {\text{V}}_{\text{1}}{\text{=V}}_{\text{2}}\text{=V}\\ \text{from}\text{the}\text{above}\text{equation}\text{we}\text{can}\text{write as }\\ \frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{= }\frac{{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The change in volume when the temperature of a gas is increased at constant pressure has to be calculated.

Concept Introduction:

STP in chemistry is the short form for Standard Temperature and Pressure.  STP most usually is used when performing calculations on gases, such as gas density.

Use the combined gas law

$\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}$

$\begin{array}{l}\text{where,}{\text{P}}_{\text{1}}\text{,}{\text{V}}_{\text{1}}\text{ and}{\text{T}}_{\text{1}}\text{are}\text{the}\text{initial}\text{}\text{pressure,}\text{volume,and}\text{temperature}\text{.}\\ {\text{P}}_{\text{2}}{\text{,V}}_{\text{2}}\text{and}{\text{T}}_{\text{2}}\text{are}\text{the}\text{final}\text{pressure,}\text{volume,}\text{and}\text{temperature}\text{.}\\ \text{pressure}\text{remains}\text{same}\text{. so,}\\ {\text{V}}_{\text{1}}{\text{=V}}_{\text{2}}\text{=V}\\ \text{from}\text{the}\text{above}\text{equation}\text{we}\text{can}\text{write as }\\ \frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{= }\frac{{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\end{array}$