Chapter 13.1, Problem 6RQ

Interpretation Introduction

Interpretation:

The new volume of the gas have to be calculated.

Concept Introduction:

Charles’s law: The volume of a given amount of gas is proportional to its temperature (in K) of the gas at constant pressure.

The relationship between volume of a gas and temperature is expressed as-

  $\begin{array}{l}V\propto T\\ \mathrm{or},\frac{V}{T}=Constant\end{array}$ When P is constant.

Charles’s law can also be expressed in mathematical form as-

  $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

In the above equation, V₁ is initial volume and T₁ is initial temperature, while V₂ is final or new volume and T₂ is final or new temperature.

Answer to Problem 6RQ

The new volume of the gas is 578 mL.

Explanation of Solution

Data given: V₁ =675 mL

T₁ = 75 °C = (273 + 75) K = 348 K

T₂ = 25 °C = (273 + 25) K = 298 K

  $\begin{array}{l}\frac{V_1}{T_1}=\frac{V_2}{T_2}\\ V_2=\frac{V_1\times T_2}{T_1}=\frac{675mL\times 298K}{348K}=578mL\end{array}$