Chapter U3, Problem C10.4RE

(a)

Interpretation Introduction

Interpretation: Proportionality constant value k at the start of the day needs to be explained.

Concept Introduction: According to Charles’ law, for a sample of gas, at a constant temperature, the ratio of the volume of gas to its Kelvin temperature is constant.

   $k\text{=}\frac{V}{T}$

The proportionality constant, k, is different for each different amount of gas.

Answer to Problem C10.4RE

Proportionality constant value at the start of the day is$10\text{mL/K}$ .

Explanation of Solution

Calculate the proportionality constant for a particular gas sample take$T_1$ is the temperature and$V_1$ is the volume of the gas. Therefore, k becomes

   $k=\frac{V_1}{T_1}$

Here,

  $T_1=23°\text{C}$

    $\text{=23+273K}$

    $\text{=296}\text{K}$

  $V_1=3\text{L}$

Therefore,

   $k=\frac{V_1}{T_1}$

    $=\frac{3\text{L}}{296\text{K}}$

    $=0.010\text{L/K}$

    $\text{=0}\text{.010}\times \text{1000}\text{mL/K}$

   $k=10\text{mL/K}$

   $k=10\text{mL/K}$

Therefore, k value is$10\text{mL/K}$ .

(b)

Interpretation Introduction

Interpretation: Proportionality constant value k at the end of the day needs to be explained.

Concept Introduction: According to Charles’ law, for a sample of gas, at a constant temperature, the ratio of the volume of gas to its Kelvin temperature is constant.

   $k\text{=}\frac{V}{T}$

The proportionality constant, k, is different for each different amount of gas.

Answer to Problem C10.4RE

Proportionality constant value at the end of the day is$10\text{mL/K}$ .

Explanation of Solution

As the other conditions such as pressure and amount of gas remain unchanged, the value of k is constant for any volume and temperature. Therefore, the proportionality constant at end of the day is the same as the start of the day.

Calculate the proportionality constant for a particular gas sample take${\text{T}}_{\text{1}}$ is the temperature and${\text{V}}_{\text{1}}$ is the volume of the gas. Therefore, k becomes

   $k=\frac{V_1}{T_1}$

Here,

  $T_1=23°\text{C}$

    $\text{=23+273K}$

    $\text{=296}\text{K}$

  $V_1=3\text{L}$

Therefore,

   $k=\frac{V_1}{T_1}$

    $=\frac{3\text{L}}{296\text{K}}$

    $=0.010\text{L/K}$

    $\text{=0}\text{.010}\times \text{1000}\text{mL/K}k=10\text{mL/K}$

Thus,

   $k=10\text{mL/K}$

Therefore, k value is$10\text{mL/K}$ .

(c)

Interpretation Introduction

Interpretation: The volume of the balloon at temperature$\text{10}°\text{C}$ needs to be explained.

Concept Introduction: According to Charles’ law, for a sample of gas, at a constant temperature, the ratio of the volume of gas to its Kelvin temperature is constant.

   $k\text{=}\frac{V}{T}$

The proportionality constant, k, is different for each different amount of gas.

Answer to Problem C10.4RE

The volume of the balloon at the temperature$\text{10}°\text{C}$ is$28.3\text{L}$ .

Explanation of Solution

Proportionality constant, k, value is$0.010\text{L/K}$ .

  $T_1=10°\text{C}$

    $\text{=10+273K}$

    $\text{=283}\text{K}$

Substitute above values in Charles law expression as follows:

   $k=\frac{V_1}{T_1}$

  $V_1\text{=283}\text{K}\times 0.010\text{L/K}$

    $\text{=}28.3\text{L}$

Therefore, volume is$28.3\text{L}$ .