Ideal Gas Law:
PV = N k T --->
N = PV/(k T)
N = 1.387*10^(29)
mass of one helium atom is approximately
4*u = 6.642*10^27 kg
4u*N = 921 kg
A Goodyear blimp typically contains 4770 m3 of helium (He) at an absolute pressure of 1.10 x 105 Pa. The temperature of the helium is 274 K. What is the mass (in kg) of the helium in the blimp?
I would help but my phisycs class is not as far as you yet.
Thanks
No problem! I can help you solve this problem step-by-step.
To find the mass of the helium in the blimp, we can use the Ideal Gas Law equation you provided:
PV = NkT
Where:
P = Pressure of the gas in Pascals (Pa)
V = Volume of the gas in cubic meters (m^3)
N = Number of gas molecules
k = Boltzmann constant (1.38 × 10^-23 J/K)
T = Temperature of the gas in Kelvin (K)
We are given:
P = 1.10 x 10^5 Pa
V = 4770 m^3
T = 274 K
First, let's calculate the value of N using the equation you provided:
N = PV / (kT)
N = (1.10 x 10^5 Pa) * (4770 m^3) / ((1.38 × 10^-23 J/K) * 274 K)
N = 1.913 x 10^27 molecules of helium
Now, we know the number of helium molecules in the blimp, but we need to find the mass of the helium. We know that the mass of one helium atom is approximately 4u, where 1u is the atomic mass unit (1u = 1.66 x 10^-27 kg).
The mass of one helium atom is approximately 4u = 6.642 x 10^-27 kg.
Now, we can find the total mass of the helium in the blimp:
Total mass = mass of one helium atom * number of helium atoms
= 6.642 x 10^-27 kg * 1.913 x 10^27 molecules
≈ 1.267 kg
Therefore, the mass of the helium in the blimp is approximately 1.267 kg.