# Question 2a2bf

Oct 25, 2015

Here's what I got.

#### Explanation:

The question is very vague, but I assume that you want to know what will happen to the pressure if you decrease temperature from ${100}^{\circ} \text{C}$ to ${85}^{\circ} \text{C}$.

Since no mention of volume or number of moles of gas was made, you can assume that they are kept constant.

In this case, pressure and temperature have a direct relationship - this is known as Gay Lussac's Law.

When volume and number of moles of gas are kept constant, increasing temperature will cause the pressure to increase as well.

Likewise, decreasing temperature, like you have in your example, will cause the pressure to decrease as well.

Mathematically, Gay Lussac's Law can be written like this

$\textcolor{b l u e}{{P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}} \text{ }$, where

${P}_{1}$, ${T}_{1}$ - the pressure and temperature of the gas at an initial state.
${P}_{2}$, ${T}_{2}$ - the pressure and temperature of the gas at a final state.

Rearrange this equation and solve for ${P}_{2}$ - do not forget that the temperature must be expressed in Kelvin!

${P}_{2} = {T}_{2} / {T}_{1} \cdot {P}_{1}$

P_2 = ( (273.15 + 85)color(red)(cancel(color(black)("K"))))/((273.15 + 100)color(red)(cancel(color(black)("K")))) * "760 torr" = color(green)("730 torr")

Indeed, the pressure decreased sa a result of a decrease in temperature.

Check out this very cool video on Gay Lussac's Law!

Oct 25, 2015

I'm guessing you needed the final pressure. Answer is 729.45 Torr or 0.96 atm.

#### Explanation:

Gay Lussac's Law is expressed by the following formula:

${P}_{1} {T}_{2}$ = ${P}_{2} {T}_{1}$ where T should be in Kelvin (K)

This means, that at a constant mass and volume, the gas's temperature should increase when its pressure increases.

So substituting the values given in the formula, we have:

${P}_{1}$ = 760 torr
${T}_{1}$ = 100 deg C + 273.15 = 373.15 K
${T}_{2}$ = 85 deg C + 273.15 = 358.15 K
${P}_{2}$ = ?

${P}_{1} {T}_{2}$ = ${P}_{2} {T}_{1}$
(760 torr) (358.15 K) = ${P}_{2}$(373.15 K)

Thus, ${P}_{2}$ = 729.45 Torr

or

${P}_{2}$ = $729.45 \cancel{\text{torr}}$ x "1 atm"/(760 cancel "torr")# = 0.96 atm