What temperature (in Celsius) did an ideal gas shift to if it was initially at -22.5 Celsius at 4.62 atm and 35.0 L and the pressure was changed to 2.50 atm and the volume changed to 50.0 L?

Answer 1

Answer: about -79.4 degrees C (193.8 K).

The amount of gas never changes here, so n and R stay constant and the quantity PV/T is the same before and after. That is the combined gas law:

P1 * V1 / T1  =  P2 * V2 / T2

Solve for T2:

T2 = T1 * (P2 * V2) / (P1 * V1)

Step 1 - convert to Kelvin FIRST. This is the step that sinks most attempts at this problem. The gas laws are built on absolute temperature. A Celsius value has an arbitrary zero, and a ratio of arbitrary-zero numbers is meaningless - at 0 degrees C you would be dividing by zero, and a negative Celsius value flips the sign of the whole answer.

T1 = -22.5 + 273.15 = 250.65 K

Step 2 - evaluate each PV product.

P1 * V1 = 4.62 atm * 35.0 L = 161.7 atm.L
P2 * V2 = 2.50 atm * 50.0 L = 125.0 atm.L

ratio = 125.0 / 161.7 = 0.7730

Step 3 - scale the absolute temperature by that ratio, then convert back.

T2 = 250.65 K * 0.7730 = 193.76 K

T2 = 193.76 - 273.15 = -79.39
   = -79.4 degrees C

The free sanity check - do this before you trust the arithmetic. Look at which way each change pushes the temperature, and multiply the two factors:

pressure factor:  2.50 / 4.62 = 0.541  (cools)
volume factor:   50.0 / 35.0 = 1.429  (warms)

net:  0.541 * 1.429 = 0.773  ->  net cooling

The net factor is less than 1, so T2 has to land BELOW 250.65 K. If your answer came out warmer than -22.5 degrees C, you either inverted the ratio or skipped the Kelvin conversion. Inverting it gives 250.65 * 1.2936 = 324.2 K = +51.1 degrees C - a perfectly reasonable-looking number, which is exactly what makes that error dangerous.

A sig-fig trap specific to this problem. Do not round 193.76 K to three significant figures (194 K) before subtracting - that would hand you -79.2 degrees C. Rounding rules change when you switch operations: multiplication and division follow significant figures, but addition and subtraction follow DECIMAL PLACES. Both 193.76 and 273.15 carry two decimals, so the difference keeps two decimals (-79.39), and only then do you report -79.4 to match the three-sig-fig inputs. Carry the extra digits and round once, at the very end.

Also worth knowing: using 273 instead of 273.15 gives 250.5 * 0.7730 = 193.64 K = -79.36, which still rounds to -79.4 degrees C. That shortcut happens to be harmless here, but it will not always be - it costs you nothing to use 273.15.

A physical anchor for the answer. -79.4 degrees C is essentially the temperature of dry ice, which sublimes at -78.5 degrees C at 1 atm. That is a useful landmark to keep in your head: if a gas-law answer lands near -79, it is a believable laboratory temperature. It is also the same physics behind a compressed-gas can going cold in your hand - the gas inside expands and drops in pressure, and its temperature falls with it.