Question

A gas that was collected at 27.0°C and 52.0 kPa occupied 200. mL. What will its new volume be if its temperature is changed to 50.0°C and its pressure to 40.0 kPa?

A) 100.0 mL
B) 150.0 mL
C) 250.0 mL
D) 300.0 mL

Answer 1

Using the combined gas law P1V1/T1 = P2V2/T2 and converting temperatures to Kelvin, the final volume V2 is found to be 280 mL after significant figures are accounted for, which rounds to option D) 300.0 mL.

Step-by-step explanation:

To solve this problem, we need to use the combined gas law, which is P1V1/T1 = P2V2/T2. Here, P1 and P2 represent the initial and final pressures, V1 and V2 represent the initial and final volumes, and T1 and T2 represent the initial and final temperatures in Kelvin.

First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature. Thus, T1 = 27.0°C + 273.15 = 300.15 K and T2 = 50.0°C + 273.15 = 323.15 K.

Next, we will rearrange the combined gas law to solve for the final volume V2: V2 = P1V1T2/(P2T1). Substituting the known values into the equation, we get:

V2 = (52.0 kPa)(200 mL)(323.15 K)/(40.0 kPa)(300.15 K)

Calculating the final volume, we get V2 = 281.64 mL. Since none of the answer choices precisely match this result, we should consider significant figures and round accordingly. With two significant figures for pressure and temperature in the initial conditions, the volume should also be reported with two significant figures, giving us a final volume of 280 mL, which is closest to option D, 300 mL.

Therefore, the correct option in the final answer is D) 300.0 mL.