Question

При какой температуре азот масса которого 1г и объем 831л,будет иметь давление 1кПа?

Answer 1

PV=mRT/μ
T=PVμ/Rm
P=1кПа=1000Па

m=1г
R=8.31=const

V=831л
μ=14 (азот)

T=1400000К

Answer 2

Calculating the Temperature of Nitrogen

To calculate the temperature at which nitrogen, with a mass of 1g and a volume of 831L, will have a pressure of 1kPa, we can use the ideal gas law formula:

PV = nRT

Where: - P = pressure - V = volume - n = number of moles - R = ideal gas constant - T = temperature in Kelvin

First, we need to calculate the number of moles of nitrogen using the given mass and molar mass of nitrogen.

The molar mass of nitrogen (N2) is approximately 28 g/mol.

Using the formula:

n = mass / molar mass

We can calculate the number of moles:

n = 1g / 28 g/mol = 0.0357 moles

Now, we can rearrange the ideal gas law formula to solve for temperature:

T = (P * V) / (n * R)

Given: - P = 1 kPa - V = 831 L - n = 0.0357 moles - R = 8.314 J/(mol*K) (ideal gas constant)

Substitute the values into the formula:

T = (1 kPa * 831 L) / (0.0357 moles * 8.314 J/(mol*K))

Calculating the temperature:

T ≈ 280 K

So, the temperature of nitrogen, with a mass of 1g and a volume of 831L, at which it will have a pressure of 1kPa, is approximately 280 Kelvin.