A sample of Nitrogen occupies a volume of 1dm3 at 500k and 1.01×10-5 newton per meter what will it volume
be at 600k and 1.01×10-5 newton per meter?
To solve this problem, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (1.01×10-5 N/m^2)
V = volume (1 dm3 = 0.001 m^3)
n = number of moles
R = ideal gas constant (8.31 J/(mol*K))
T = temperature (in Kelvin, 500 K and 600 K)
First, we need to calculate the number of moles of Nitrogen in the initial state:
PV = nRT
n = PV / RT
n = (1.01×10-5 N/m^2 * 0.001 m^3) / (8.31 J/(mol*K) * 500 K)
n = 2.42×10-8 moles
Now, we can use the number of moles to find the final volume at 600 K:
V = nRT / P
V = (2.42×10-8 moles * 8.31 J/(mol*K) * 600 K) / 1.01×10-5 N/m^2
V = 0.001 m^3
Therefore, the final volume of Nitrogen at 600 K and 1.01×10-5 N/m^2 would be 0.001 m^3 (1 dm3).