Chapter U3.6, Problem 6E

Interpretation Introduction

Interpretation: The temperature to which the gas must be cooled for the volume to change to 1.0L is to be calculated.

Concept Introduction: According to Charles’s law, if pressure and the number of particles of gas remains the same, then the volume is proportional to the Kelvin temperature.

  $V=kT$

Answer to Problem 6E

The temperatureto which the gas has to be cooled is $126{}^{\text{o}}\text{C}$ .

The following graph shows the change in volume as the temperature is decreased.

  

Explanation of Solution

Given information:

A sample of gas in a cylinder has a volume of 2.0L at 20⁰C. It is cooled down to have a volume of 1.0L.

As the temperature decreases, the volume of gas also decreases. It is known that volume is directly proportional to temperature.

First, change the Celsius temperature to Kelvin

So, $20+273=293\text{ K}$

Using the formula,

  $k=\frac{V}{T}$

One can first find the value of K;

  $k=\frac{2000\text{ mL}}{293\text{ K}}=6.8\text{ mL/K}$

Now solving for the temperature for the volume of 1.0L;

  $T=\frac{1000\text{ mL}}{6.8\text{ mL/K}}=147\text{ K}$

Changing Kelvin temperature to degree Celsius

  $147\text{ K=147-273=-126}{\text{.0 }}^{\text{o}}\text{C}$

One can find out the temperature using the graph as well.

  

First, plot a point with initial conditions. One can draw a line through this point and the origin since at 0K, the volume is theoretically 0mL

Then use the line to find the temperature when the y-value is 1000L.

When the volume is reduced to 1.0L, the temperature will be $\text{-126}{\text{.0 }}^{\text{o}}\text{C}$ .