Question #dfea5

1 Answer
Mar 30, 2016


#"25 atm"#


Notice that the problem doesn't provide information about the number of moles of gas present in the balloon and about the balloon's volume, which can only mean that you are supposed to assume that they remain constant.

When number of moles of gas and volume are kept constant, temperature and pressure have a direct relationship #-># this is known as Gay Lussac's Law.

Simply put, when temperature increases, pressure increases as well, and when temperature decreases, pressure decreases as well.

This happens because gas pressure depends on the frequency and force of the collisions that occur between the gas molecules and the walls of the balloon.

When you decrease the temperature of the gas, you're essentially decreasing the average kinetic speed of its molecules.

The molecules will now hit the walls of the container less frequent and with less force, which is why the pressure of the gas will decrease.

Mathematically, this is expressed as

#color(blue)(|bar(ul(color(white)(a/a)P_1/T_1 = P_2/T_2color(white)(a/a)|)))" "#, 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 final state

Before plugging in your values, make sure that you convert the temperature from degrees Celsius to Kelvin by using the conversion factor

#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 #P_2#

#P_1/T_1 = P_2/T_2 implies P_2 = T_2/T_1 * P_1#

In your case, you will have

#P_2 = ([273.15 -(15.0)]color(red)(cancel(color(black)("K"))))/((273.15 + 32)color(red)(cancel(color(black)("K")))) * "29.6 atm" = "25.04 atm"#

Rounded to two sig figs, the number of sig figs you have for the initial temperature of the gas, the answer will be

#P_2 = color(green)(|bar(ul(color(white)(a/a)"25 atm"color(white)(a/a)|)))#

Notice that the decrease in temperature resulted in a decrease in pressure.