Chapter 21, Problem 91A

Interpretation Introduction

Interpretation: The new volume of the oxygen gas needs to be determined if the temperature is increased by ${\text{60 }}^{\text{o}}\text{C}$ at constant pressure. The initial volume and temperature of the oxygen gas is 425 mL and ${\text{30 }}^{\text{o}}\text{C}$ respectively.

Concept Introduction: The relation between pressure, volume and, temperature of a gas can be obtained from ideal gas law. This is represented as follows:

  $PV=nRT$

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature.

Explanation of Solution

For a gas at constant pressure, the relation between initial and final volume and temperature can be represented s follows:

  $P{V}_1=nR{T}_1$   ....... (1)

Here, sample is same thus, in both cases number of moles will be same and R is universal gas constant so that will also be same. Also, pressure is constant so it will also be same. The final volume and temperature can be represented as follows:

  $P{V}_2=nR{T}_2$   ....... (2)

Dividing equation (1) and (2),

  $\begin{array}{l}\frac{P{V}_1}{P{V}_2}=\frac{nR{T}_1}{nR{T}_2}\\ \frac{V_1}{V_2}=\frac{T_1}{T_2}\end{array}$

Here, temperature must be in Kelvin.

The initial volume is 425 mL. The initial and final temperatures in K will be:

  $\begin{array}{l}{T}_1=30{}^{\text{o}}\text{C}+\text{273}\text{.15 K}=303.15\text{ K}\\ T_2=60{}^{\text{o}}\text{C}+\text{273}\text{.15 K}=333.15\text{ K}\end{array}$

Substitute the values,

  $\frac{425\text{ mL}}{V_2}=\frac{303.15\text{ K}}{333.15\text{ K}}$

Now,

  $\begin{array}{c}{V}_2=\frac{425\text{ mL}}{303.15\text{ K}}\times 333.15\text{ K}\\ =467.06\text{ mL}\end{array}$

Thus, new volume of the oxygen gas will be 467.06 mL.