# Question 674f8

Jul 27, 2017

$- {40}^{\circ} \text{C}$

#### Explanation:

Your tool of choice here will be the equation that describes Charles' Law

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}}}}$

Here

• ${V}_{1}$ and ${T}_{1}$ represent the volume and the absolute temperature of the gas at an initial state
• ${V}_{2}$ and ${T}_{2}$ represent the volume and the absolute temperature of the gas at a final state

This equation shows that when the pressure of a gas and the number of moles of gas present in the sample remain constant, the volume of the gas and its temperature have a direct relationship.

In other words, when the pressure and the number of moles of gas are constant, an increase in the temperature of the gas will cause its volume to increase as well.

Similarly, a decrease in the temperature of the gas will cause its volume to decrease as well.

In your case, you have a gas at normal temperature, which is equal to ${20}^{\circ} \text{C}$ or

${20}^{\circ} \text{C" + 273.15 = "293.15 K}$

You know that the volume of the gas must decrease by 20% of its initial value, so

${V}_{2} = {V}_{1} - \frac{20}{100} \cdot {V}_{1} = 0.8 \cdot {V}_{1}$

Right from the start, you can say that because the volume is decreasing, you must have

${T}_{2} < \text{293.15 K}$

Rearrange the equation to solve for ${T}_{2}$

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2} \implies {T}_{2} = {V}_{2} / {V}_{1} \cdot {T}_{1}$

Plug in your values to find

${T}_{2} = \frac{0.8 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{1}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{1}}}}} \cdot \text{293.15 K" = "234.52 K}$

This will be equivalent to

t[""^@"C"] = "234.52 K" - 273.15 = color(darkgreen)(ul(color(black)(-40^@"C")))#

The answer must be rounded to one significant figure, the number of sig figs you have for the percent decrease.

Notice that you don't really need to know that the gas occupies ${\text{500 cm}}^{3}$ at normal temperature.