the largest a party balloon can get before bursting is 8.23L at 25 degrees celsius and 1.00 atm. supposing you fill the balloon only with oxygen gas, how many grams of oxygen can be added to the balloon before it pops? assume you have not added any gas to the baloon at all (so essentially it is flat)
a.) 10.8 g O2
b.) 0.337 g O2
c.) 4.01 g O2
d.) 128 g O2
e.) 2.97 g O2
after some calculations i assume the answer is D, but my friend keeps saying it is B
I think both of you are incorrect. Use PV = nRT and solve for n, then grams = n x molar mass O2.
Well, let's figure this out. Given the volume, temperature, and pressure of the balloon, we can use the ideal gas law to calculate the amount of oxygen it can hold before bursting. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, let's convert the volume from liters to cubic meters (m³) because the ideal gas constant has units of Pa·m³ / (mol·K). So, 8.23 L is equal to 0.00823 m³.
Next, let's convert the temperature from degrees Celsius to Kelvin. Adding 273.15 to 25 degrees Celsius gives us 298.15 K.
Now, we can rearrange the ideal gas law equation to solve for n (the number of moles): n = PV / RT.
We know that the pressure is 1.00 atm, which is approximately equal to 101325 Pa. The ideal gas constant R is 8.314 J / (mol·K).
Now we can plug in the values into the equation: n = (101325 Pa) x (0.00823 m³) / ((8.314 J/(mol·K)) x (298.15 K)).
After doing the calculations, we find that n ≈ 0.337 mol.
Since we want to find the mass of oxygen in grams, we need to multiply the number of moles by the molar mass of oxygen, which is approximately 32 g/mol (16 g/mol for oxygen, times 2 because it's O₂).
So, the mass of oxygen that can be added to the balloon before it pops is approximately 0.337 mol x 32 g/mol ≈ 10.8 g of oxygen.
Therefore, the correct answer is a.) 10.8 g O₂. Your friend was right! Enjoy the burst of knowledge!
To determine the correct answer, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = ideal gas constant
T = temperature
First, let's convert the given volume to liters using the conversion factor: 1 L = 1000 cm³
8.23 L * (1000 cm³/1 L) = 8230 cm³
Next, let's convert the temperature from Celsius to Kelvin by adding 273:
25°C + 273 = 298 K
Now, we can rearrange the ideal gas law equation to solve for the moles of gas:
n = PV / RT
P = 1.00 atm
V = 8230 cm³/1000 = 8.23 L
R = 0.0821 L.atm/(mol.K)
T = 298 K
n = (1.00 atm) * (8.23 L) / (0.0821 L.atm/(mol.K)) * (298 K)
n = 0.319 mol
Since the molar mass of oxygen is approximately 32 g/mol, we can calculate the mass of oxygen:
Mass = n * molar mass
Mass = 0.319 mol * 32 g/mol
Mass = 10.208 g
Therefore, the correct answer is not listed among the given options. The approximate mass of oxygen that can be added to the balloon before it bursts is 10.21 g, which is not one of the given options.
To solve this question, we can utilize the ideal gas law equation: PV = nRT.
First, let's determine the initial conditions of the balloon when it's flat. We are given:
- Volume (V) = 0 since the balloon is flat.
- Temperature (T) = 25 degrees Celsius, which we need to convert to Kelvin by adding 273.15 K.
- Pressure (P) = 1.00 atm.
Next, we need to find the final volume (V_final) of the balloon when it is filled with gas before bursting. The given final volume is 8.23 L.
Using the ideal gas law equation, we can rearrange it to solve for the amount of moles (n) of oxygen gas that can be added to the balloon:
n = PV / RT
Let's substitute the values into the equation:
n = (1.00 atm) * (8.23 L) / [0.0821 L·atm/(mol·K) * (25 + 273.15) K]
Calculating this expression gives us the number of moles of oxygen gas that can be added to the balloon.
To convert moles (n) to grams of oxygen (O2), we need to use the molar mass of oxygen. The molar mass of O2 is approximately 32 g/mol.
Final step:
mass = moles * molar mass
mass = n * (32 g/mol)
By plugging in the number of moles calculated earlier into this formula, we can determine the final mass of the oxygen gas that can be added to the balloon before it pops.
Calculating this expression gives us the answer to the question.
Comparing this result with the given options, it appears that option D (128 g O2) is the correct answer.
Thus, the correct answer is: d.) 128 g O2.