A gas sample was originally in a 2.50 × 10² mL container at 25 °C.The gas was transferred to an evacuated 5.00 × 10² mL container,and the pressure was the same in both containers.What is the temperature of the gas in the larger container? A)596 °C B)323 °C C)50 °C D)13 °C E)-124 °C

Question

Quizplus · Accepted Answer

According to the ideal gas law (PV = nRT), the pressure (P), volume (V) and number of moles (n) of a gas are directly proportional to the absolute temperature (T). So, if the pressure and number of moles are kept constant, the ratio of the volumes is equal to the ratio of the temperatures.

Therefore, the temperature of the gas in the larger container can be calculated as:

T2 = (V2/V1) x T1

where T1 = 25°C + 273.15 = 298.15 K (convert to Kelvin)
V1 = 2.50 × 1012 mL = 2.50 × 10-3 L
V2 = 5.00 × 1012 mL = 5.00 × 10-3 L

T2 = (5.00 × 10-3 L /2.50 × 10-3 L) x 298.15 K = 596 K

So, the temperature of the gas in the larger container is 596 K - 273.15 = 322.85 ≈ 323°C, which is answer choice B.

A Gas Sample Was Originally in a 2