How Does Compressing an Ideal Gas Affect Its Temperature?

[Bgerst103]

Homework Statement

A cylinder fitted with a movable piston contains ideal gas at 27C, pressure .5 x 10^5 Pa, and volume 1.25 m^3. What will be the final temperature if the gas is compressed to .8 m^3 and the pressure rises to .82 x 10^5 Pa.

A) 125 C
B) 154 C
C) 246 C
D) 67.7 C
E) 41.8 C

Homework Equations

P1V1/T1=P2V2/T2

The Attempt at a Solution

I thought this problem was going to be pretty simple. (.5 x 10^5 Pa x 1.25 m^3)/27 = (.82 x 10^5 Pa x .8 m^3)/T2. I keep getting T2 is equal to 28.3392. Is there something I'm missing? It seems like the pressure/volume ratio doesn't change that drastically so the temperatures would be close but the closest answer is at ~15 C increase. Any help is appreciated.

 

[Chestermiller]

You have to use absolute temperatures.

Chet

 

[Bgerst103]

You have to use absolute temperatures.

Chet

Got it. I thought since it was ratios the units wouldn't affect the final outcome but I was wrong.

 

[dauto]

Only when you're talking about temperature DIFFERENCES - not ratios - you may replace Kelvins with Celsius.

 

[HalfLostAndSearching]

Your attempt at the solution is correct, but the final temperature you calculated (28.3392 C) is in Kelvin, not Celsius. To convert from Kelvin to Celsius, you need to subtract 273 from the temperature. Therefore, the final temperature in Celsius would be 28.3392 - 273 = -244.6608 C. This is the closest to the answer of 246 C, so your solution is correct.