Question

A gas sample originally occupies 436 mL at 24 C. When the volume is expanded to 612 mL, and the temperature is increased to 97 C, the pressure becomes 526 mm Hg. What was the original pressure?

Answer 1

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

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

Where: P1 = Initial pressure (unknown) V1 = Initial volume T1 = Initial temperature P2 = Final pressure V2 = Final volume T2 = Final temperature

Given: V1 = 436 mL T1 = 24 °C = 24 + 273.15 K = 297.15 K V2 = 612 mL T2 = 97 °C = 97 + 273.15 K = 370.15 K P2 = 526 mm Hg

Substituting these values into the combined gas law equation, we have:

(P1 * 436 mL) / (297.15 K) = (526 mm Hg * 612 mL) / (370.15 K)

Now, we can solve for P1 by rearranging the equation:

P1 = (526 mm Hg * 612 mL * 297.15 K) / (436 mL * 370.15 K)

P1 ≈ 361.6 mm Hg

Therefore, the original pressure is approximately 361.6 mm Hg.