A 3.50L balloon contains 0.178 mol of gas. If 0.094 mol of gas is added to the balloon, what will be the new volume?
5.35 L
Well, if the balloon was a person, it would probably say, "Oh no, I ate too many beans!" But since it's a balloon, it won't have any hilarious digestive consequences. So, let's do some math instead. We started with a 3.50L balloon containing 0.178 mol of gas. If we add 0.094 mol of gas to it, the total number of moles becomes 0.178 mol + 0.094 mol = 0.272 mol. Now it's time to find the new volume. Drum roll, please! (Insert drum roll sound here) The new volume will be 0.272 mol divided by the original number of moles (0.178 mol) multiplied by the original volume (3.50 L). So, the new volume of the balloon will be approximately 5.40 L. Voila! We've inflated the balloon with some math magic.
new mols = 0.178 + 0.094 = ?
Then PV = nRT. Remember T must be in kelvin.
To determine the new volume of the balloon after 0.094 mol of gas is added, we can use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
In this case, we need to assume constant temperature and pressure since the problem does not provide any information about changes in those parameters. Thus, we can rewrite the ideal gas law equation as:
V1 = (n1 * R * T) / P
Where V1 is the initial volume of the balloon, n1 is the initial number of moles of gas, R is the ideal gas constant, and P is the pressure.
Now, we can calculate the initial volume of the balloon using the given information:
V1 = (0.178 mol * R * T) / P
Since the initial volume is given as 3.50L, we can set up an equation:
3.50L = (0.178 mol * R * T) / P
Next, we'll solve for the value of R * T / P, which is a constant:
R * T / P = (3.50L * P) / (0.178 mol)
Now that we have the value for R * T / P, we can use this constant to calculate the new volume when 0.094 mol of gas is added. Let's call this new volume V2:
V2 = (n1 + n2) * (R * T) / P
where n2 is the quantity of gas added, and V2 is the new volume.
Plugging in the values, we get:
V2 = (0.178 mol + 0.094 mol) * (R * T) / P
V2 = 0.272 mol * (R * T) / P
Since we know the value of R * T / P from the previous calculation, we can substitute it in:
V2 = 0.272 mol * (3.50L * P) / (0.178 mol)
Simplifying further, we get:
V2 = 5.382L + P
Therefore, the new volume will be approximately 5.382L plus the pressure (P).