Useful Constants: 1 Torr = 133.32 Pa; R = 8.3145 J/mol·K
You have a thin metal sphere of unknown volume that contains helium gas at low pressure. You put the entire metal sphere into a bath of liquid nitrogen, and a pressure gauge on the sphere indicates a pressure of 288 torr.
Given that the volume of the metal sphere, pressure gauge, etc., was 367 mL, how many moles of helium gas would you have?
temp of liquid nitrogen=77.2K
I used pV=nRT
Which would be 288*367=n*8.3145*77.2
I got n=641.8811 moles
But the answer is wrong can anyone help?
anyone????
To find the number of moles of helium gas in the metal sphere, you correctly used the ideal gas law equation, PV = nRT. However, there is a small mistake in your calculation.
The equation can be rearranged as follows:
n = (PV) / (RT)
You correctly substituted the values:
P = 288 torr
V = 367 mL = 0.367 L (since 1 L = 1000 mL)
R = 8.3145 J/mol·K
T = 77.2 K
Now let's plug in these values into the equation:
n = (288 torr * 0.367 L) / (8.3145 J/mol·K * 77.2 K)
We need to convert the pressure in torr to pascals (Pa) to match the units of the gas constant (R). Recall that 1 Torr = 133.32 Pa.
So, let's convert the pressure:
P = 288 torr * 133.32 Pa/Torr = 38399.36 Pa
Now plug in the converted pressure value:
n = (38399.36 Pa * 0.367 L) / (8.3145 J/mol·K * 77.2 K)
Calculating this, you will find:
n ≈ 22.6 moles
Therefore, the correct number of moles of helium gas in the metal sphere is approximately 22.6 moles.