A balloon containing 0.600 mole of SO2 gas has a volume of 275 mL at a certain temperature and pressure. What volume, in milliliters, would a 0.765 mole sample of SO2 gas occupy at the same temperature and pressure conditions?

The easy way to do this is to recognize that volume is proportional to mols of gas at the same T and P or

(n1/v1) = (n2/v2) but an equally easy, but more work is to assume some value for T (any convenient numbers will work) and use PV = nRT, substitute the numbers and solve for P. Then using that value for P and the assumed value for T, use PV = nRT, substitute those numbers and solve for the new volume. You should get the same answer either way. Post your work if you get stuck.

To solve this problem, we can use the ideal gas law equation:

PV = nRT,

where:
P = pressure (constant)
V = volume of gas
n = number of moles of gas
R = ideal gas constant
T = temperature (constant)

We are given that the temperature and pressure are constant for both scenarios, so we can rewrite the equation as:

V1 / n1 = V2 / n2,

where:
V1 = initial volume (275 mL)
n1 = initial number of moles (0.600 moles)
V2 = final volume (unknown)
n2 = final number of moles (0.765 moles)

Now, let's rearrange the equation to solve for V2:

V2 = (V1 * n2) / n1.

Substituting the given values into the equation:

V2 = (275 mL * 0.765 moles) / 0.600 moles.

V2 = (209.875 mL) / 0.6.

V2 ≈ 349.79 mL.

Therefore, a 0.765 mole sample of SO2 gas would occupy approximately 349.79 milliliters at the same temperature and pressure conditions.