Question #68da5

1 Answer
Mar 18, 2016

Answer:

#"250 K"#

Explanation:

Your tool of choice here will be the combined gas law equation, which looks like this

#color(blue)(|bar(ul(color(white)(a/a)(P_1V_1)/T_1 = (P_2V_2)/T_2color(white)(a/a)|)))" "#, where

#P_1#, #V_1#, #T_1# - the pressure, volume, and absolute temperature of the gas at an initial state
#P_2#, #V_2#, #T_2# - the pressure, volume, and absolute temperature of the gas at a final state

Now, before you plug in your values and solve for #T_2#, try to see if you can predict what the new temperature will be relative to the initial temperature.

Start by assuming that the volume of the gas changes from #"450 mL"#" to #"350 mL"#, but that the pressure, along with the number of moles of gas, remain constant.

In this case, temperature and volume have a direct relationship known as Charles' Law.

When volume increases, temperature increases as well, and when volume decreases, temperature decreases as well.

In your case, the volume decreases, so if this was the only change in the conditions of the gas, the new temperature would be lower than the initial temperature.

Next, assume that the pressure of the gas changes from #"730 torr"# to #"770 torr"#, but that this time the volume and the number of moles of gas remain constant.

When this happens, pressure and temperature have a direct relationship known as Gay Lussac's Law.

This means that an increase in pressure would trigger an increase in temperature. If this was the only change in conditions, the new temperature would be higher than the initial temperature.

This tells you that increasing the pressure of a gas and decreasing its volume have competing effects on its temperature.

Since the decrease in volume is more significant than the increase in pressure, you can assume that the final temperature of the gas will be lower than the initial temperature.

To see which one "wins", plug in your values into the combined gas law equation. Do not forget to convert the initial temperature from degrees Celsius to Kelvin

#color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#

Rearrange the equation to solve for #T_2#

#T_2 = overbrace(P_2/P_1)^(color(blue)("increase in pressure")) * overbrace(V_2/V_1)^(color(red)("decrease in volume")) * T_1#

#T_2 = (770color(red)(cancel(color(black)("torr"))))/(730color(red)(cancel(color(black)("torr")))) * (350color(red)(cancel(color(black)("mL"))))/(450color(red)(cancel(color(black)("mL")))) * (29 + 273.15)"K"#

#T_2 = color(green)(|bar(ul(color(white)(a/a)"250 K"color(white)(a/a)|)))#

If you want, you can express this in degrees Celsius

#t_2 = 248 - 273.15 = color(green)(|bar(ul(color(white)(a/a)-23^@"C"color(white)(a/a)|)))#

Both answers are rounded to two sig figs.

As predicted, the temperature of the gas decreased because the decrease in volume was more significant than the increase in pressure.