Question

Чему равна температура азота массой 2,8 г и объемом 500 см3 при давлении 400 кПа

Answer 1

PV=m/M * RT Отсюда выразить T.
T=pVM/mR

Answer 2

Calculation of Nitrogen Temperature

To calculate the temperature of nitrogen with a mass of 2.8 g and a volume of 500 cm3 at a pressure of 400 kPa, we can use the ideal gas law equation:

PV = nRT

Where: - P is the pressure (in Pa) - V is the volume (in m3) - n is the number of moles of gas - R is the ideal gas constant (8.314 J/(mol·K)) - T is the temperature (in Kelvin)

First, let's convert the given values to the appropriate units: - Mass: 2.8 g - Volume: 500 cm3 (which is equivalent to 0.5 L) - Pressure: 400 kPa (which is equivalent to 400,000 Pa)

Next, we need to calculate the number of moles of nitrogen using its molar mass. The molar mass of nitrogen (N2) is approximately 28 g/mol.

Number of moles (n) = Mass / Molar mass

n = 2.8 g / 28 g/mol = 0.1 mol

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

T = PV / (nR)

Substituting the given values:

T = (400,000 Pa) * (0.5 L) / (0.1 mol * 8.314 J/(mol·K))

Calculating this expression gives us the temperature of nitrogen.