What would be the new temperature in Celsius of a sample of methane if 205 ml at 725 torr and 75.0 degrees C is allowed to change to 907 ml and 827 torr?
I used P1V1/T1=P2V2/T2
Plugged in the numbers (205)(725)/(348.15K)=(907)(827)/T2
When I solved for T2 I got 1.757*10^3, and I then subtracted 273.15 to get to Celsius.
Am I on the right track?
Please and thanks!
Carl, what's with all the names? Jim, Carl, George, Blaine et al. You confuse us when you do this and it makes it much harder to help you.
I didn't check the math but the procedure is correct and I see you converted all of the C to K first. That's good.
Yes, you are on the right track! To solve for the new temperature in Celsius, you correctly used the ideal gas law equation, P1V1/T1 = P2V2/T2, where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.
By substituting the given values into the equation, you obtained (205 ml)(725 torr)/(75.0 + 273.15 K) = (907 ml)(827 torr)/T2. Solving for T2, you found T2 = 1.757*10^3 K.
However, to convert this temperature from Kelvin to Celsius, you need to subtract 273.15, not add. So, subtracting 273.15 from 1757 K, you get 1,483.85 °C. Therefore, the new temperature in Celsius would be approximately 1,483.85 °C.
Great job on your calculations!