Question

A chemist produces 460 mL of oxygen gas at - 43 °C and constant tremendous pressure. To what temperature must the oxygen be warmed in order for it to have a volume of 600 mL in Celsius?

Answer 1

To solve this problem, we can use the combined gas law, which relates the initial and final conditions of a gas sample under constant pressure. The combined gas law equation is as follows:

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

Where: P1 and P2 are the initial and final pressures, which are constant in this case. V1 and V2 are the initial and final volumes. T1 and T2 are the initial and final temperatures.

Let's assign the given values to the corresponding variables:

P1 = P2 (constant pressure) V1 = 460 mL T1 = -43 °C (initial temperature) V2 = 600 mL T2 = ? (final temperature)

Now we can rearrange the equation to solve for T2:

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

Substituting the given values into the equation:

(T2) = (460 mL * -43 °C) / (600 mL)

Now we can calculate the final temperature:

T2 = (460 * -43) / 600 ≈ -33.1 °C

Therefore, the oxygen gas must be warmed to approximately -33.1 °C in order for it to have a volume of 600 mL.