Gay-Lussac’s Law
Joseph Gay-Lussac’s Law basically says that for an ideal gas if the temperature (in Kelvins) increases by a certain factor, the pressure will increase by that same factor. So if the temperature doubles, the pressure will also double.
The formula for Gay Lussac’s Law is:
P1 = P2
T1 = T2
Temperature is measured in Kelvin for this law.
Suppose you want to drive a car at top speed over the Utah sand flats. You fill the tires to their maximum pressure level, 4 atm of air at 27 degrees Celsius. You drive at 150 mph over the sand flats and the temperature of the air in the tires increases to 375 Kelvin. What is the new pressure of the air inside the tires?
P1 =
T1 =
P2 =
T2 =
Next, you solve the equation for P2 and you see that
P2 = P1T2 so you plug in the numbers: P2 = (4 atm)(375K) = 5 atm
T1 300K
Therefore, the pressure in the tires will increase to 5 atm. As the temperature rises, the particles of gas speed up and collide more often. So the pressure also rises.
Problem / Work1) A gas in a closed container has a pressure of 301 kPa at 28.0 0Celsius. If the temperature is lowered to 101 K, what will the new pressure be? (4 pts) / P1=
T1=
P2=
T2=
2) The balloons used for the Macy’s parade on Thanksgiving are supposed to be inflated to a pressure of 3 atm at 27.0 0Celsius in order to float. Assuming the ballooners used the same volume of air every year, what pressure would the balloons be inflated to if the parade happened on a really cold day, say -23.0 0Celsius? (4 pts)
3) Can you heat up a gas just by changing its pressure? Let’s see. A gas at 25.0 0Celsius and 1 atm pressure is compressed. The new pressure is 202.6 kPa. What is the new temperature? (4 pts)
4) If you had a balloon at standard pressure and 27 0C, and you released it into outer space where the temperature is 20.0 K, what would the pressure be inside the balloon? (4 pts)
5) The pressure of a gas at 27 0C is 3 atm. If the pressure drops by half, what will the new temperature of the gas be? (Give the final temp. in 0C) (5 pts)