149ml bubble of hot gases at 217 degrees celcius and 1.78 atm escapes from an active volcano. What is the temperature, in celcuis, of the gas in the bubble outside the volcano if the new volume of the bubble is 164ml and the pressure is 0.770atm?
Use (P1V1/T1) = (P2V2/T2)
Remember T must be in kelvin.
thank you so much. I was missing a step.
To solve this problem, we can use the combined gas law equation, which relates the initial and final conditions of temperature, pressure, and volume for a given amount of gas.
The combined gas law equation is as follows:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure (1.78 atm)
V1 = initial volume (149 ml)
T1 = initial temperature (217 degrees Celsius + 273.15) converted to Kelvin
P2 = final pressure (0.770 atm)
V2 = final volume (164 ml)
T2 = final temperature (what we need to find)
First, let's convert the initial temperature from Celsius to Kelvin:
T1 = 217 + 273.15 = 490.15 K
Now we can plug in the values into the equation and solve for T2:
(1.78 atm * 149 ml) / (490.15 K) = (0.770 atm * 164 ml) / (T2)
We can cross multiply and rearrange the equation:
(1.78 atm * 149 ml * T2) = (0.770 atm * 164 ml * 490.15 K)
Now we isolate T2:
T2 = (0.770 atm * 164 ml * 490.15 K) / (1.78 atm * 149 ml)
By calculating this expression, we can find T2, which represents the final temperature of the gas in Celsius.