Question

A 0.5 mole sample of argon gas occupies a volume of 500 mL at 740mm Hg pressure. At what volume will the pressure of the argon gas be 780mm Hg if the number of moles are increased to 1.2?

Answer 1

The volume of a 1.2 mole sample of argon gas, which initially has 0.5 moles at a volume of 500 mL and a pressure of 740mm Hg, will be approximately 1,138.46 mL when the pressure is increased to 780mm Hg.

Step-by-step explanation:

The student asked about the change in volume of a argon gas sample when the pressure is increased and the amount of gas (number of moles) is also increased. To solve this problem, we can use the combined gas law which is expressed as (P1 * V1) / n1 = (P2 * V2) / n2. Here, P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and n1 and n2 are the initial and final amounts of moles of the gas, respectively.

Given:
Initial conditions: P1 = 740 mmHg, V1 = 500 mL, n1 = 0.5 moles
Final conditions: P2 = 780 mmHg, n2 = 1.2 moles
We want to find the final volume (V2).

Using the combined gas law, we rearrange the formula to solve for V2:
V2 = (P1 * V1 * n2) / (P2 * n1)

Plugging in the values we have:
V2 = (740 mmHg * 500 mL * 1.2 moles) / (780 mmHg * 0.5 moles)
V2 = (444,000 mmHg*mL*moles) / (390 mmHg*moles)
V2 = 1,138.46 mL

So the final volume (V2) of the argon gas at a pressure of 780 mmHg and 1.2 moles is approximately 1,138.46 mL.