Question

You place a block of ice (mass of 3 kg) into a test chamber filled with diatomic Nitrogen gas (N₂). The chamber does not release any heat from the walls out to the environment. The measured volume of the chamber is 35 m³. You set the pressure to 7.60x10⁴ Pa.  The temperature of the Nitrogen gas is 28 °C

a.) You have the ice at the melting point so what's the temperature at equilibrium?

b.) Before you put the ice block in the chamber, calculate the v_rms of Nitrogen at 28 °C

Answer 1

a) First we calculate the process conditions in the corresponding units:

P = 76000 Pa * (1 atm / 101325 Pa) = 0.75 atm

V = 35 m3 * (1000 L / 1 m3) = 35000 L

T = 28 + 273 = 301 K

We calculate the mass of N2 in the chamber, by means of the ideal gas equation:

m N2 = P * V * MM / R * T = 0.75 atm * 35000 L * 14 g / mol / 0.082 atm * L / mol * K * 301 K = 14889.39 g

We calculate the ice mass:

m Ice = 3 Kg * (1000 g / 1 Kg) = 3000 g

To determine the final equilibrium temperature, we equate the heat equations for each sutancia:

m Ice * cp Ice * (Tf - TiH) = - m N2 * cp N2 * (Tf - TiN2)

We replace:

3000 g * 2.09 J / ºC * g * (Tf - 0ºC) = - 14889.39 g * 1.04 J / ºC * g * (Tf - 28ºC)

We cleared:

Tf / Tf - 28 = - 2.47

Clear Tf:

Tf = 69.16 / 3.47 = 19.93 ºC

b) Calculation of Vrms:

Vrms = √3 * R * T / MM = √ 3 * 0.082 * 301 K / 14 = 2.30

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