how many mols of gas will occupy 6.75 m^3 at 65°C and 3.28 x 10^5 Pa?
Convert cubic meters to cc.
6.75 m^3 x (100 cm/1m)(100 cm/1m) x (100 cm/m) = 6750000 or 6750 L.
You know that 1 mol occupies 22.4 L
6750 L x (1 mol/22.4L) = ?
p v = n R t ... n = (p v) / (R t)
convert ºC to ºK
make sure the R value has the correct units
Scott's answer is better than mine; I didn't read the entire question. My answer is good ONLY for STP and this problem is not at STP. MY eye sight let's me see a limited field and I didn't see the T and P listed. Sorry about that.
To determine the number of moles of gas that will occupy 6.75 m^3 at 65°C and 3.28 x 10^5 Pa, you can use the ideal gas law. The ideal gas law is given by the equation:
PV = nRT
Where:
- P is the pressure of the gas (in Pascals)
- V is the volume of the gas (in cubic meters)
- n is the number of moles of gas
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature of the gas (in Kelvin)
First, you need to convert the temperature from Celsius to Kelvin by adding 273.15:
65°C + 273.15 = 338.15 K
Next, rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Now, plug in the given values:
P = 3.28 x 10^5 Pa
V = 6.75 m^3
R = 8.314 J/(mol·K)
T = 338.15 K
Substitute these values into the equation to calculate the number of moles:
n = (3.28 x 10^5 Pa) * (6.75 m^3) / (8.314 J/(mol·K) * 338.15 K)
n ≈ 71.35 mol
Therefore, approximately 71.35 moles of gas will occupy a volume of 6.75 m^3 at 65°C and 3.28 x 10^5 Pa.