Question

The library of congress contains many priceless documents in plastic containers. To preserve them from oxidation, Argon gas replaces the air in the containers. One such container had 0.29 moles of Ar gas. The volume of this container 3.9L. Later in the day another attendant adds 0.37 moles of Ar to the same container. What volume will this larger amount of Ar cause the container to expand to?

Answer 1

To solve this problem, we can use the ideal gas law equation, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

Since the pressure and temperature remain constant, we can rearrange the equation to solve for volume:

V1 / V2 = n1 / n2

where V1 is the initial volume, V2 is the final volume, n1 is the initial number of moles, and n2 is the final number of moles.

Given that V1 = 3.9 L and n1 = 0.29 moles, and n2 = 0.29 moles + 0.37 moles = 0.66 moles, we can substitute these values into the equation:

3.9 / V2 = 0.29 / 0.66

To find V2, we can cross-multiply and solve:

0.29 * V2 = 3.9 * 0.66

V2 ≈ (3.9 * 0.66) / 0.29

V2 ≈ 8.93 L

Therefore, the larger amount of Argon gas will cause the container to expand to approximately 8.93 liters.