Chemistry, 06.03.2020 23:46 reggiegilbert1995

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at the constant conditions of 45∘C and 1500 kPa. A valve in the line is cracked so that air flows slowly into the tank until the pressure equals the line pressure. If the process occurs slowly enough that the temperature in the tank remains at 25∘C, how much heat is lost from the tank? Assume air to be an ideal gas for which CP=(7/2)R and CV=(5/2)R.

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Answer from: jfw0504

A heat amount of 23.461 kilojoules is lost from the tank.

According to the Equation of State for Ideal Gases, the mass of air (), in kilograms, within the tank can be found by using the following form:

$m = \frac{P\cdot V\cdot M}{R_{u}\cdot T}$ (1)

Where:

- Pressure, in kilopascals.

- Volume, in cubic meters.

- Molar mass, in kilomoles per kilogram. $R_{u}$ - Ideal gas constant, in kilopascal-cubic meters per kilomole-Kelvin

- Temperature, in Kelvin.

Heat losses () occur in a isochoric process, that is, a process at constant volume, it is caused by the heat released by the air flowing to the tank. The formula is represented by the following application of the definition of sensible heat:

$Q = \frac{5}{2}\cdot (m_2-m_1) \cdot R_{u}\cdot (T_{1}-T_{2})$ (2)

Where:

$m_{1}, m_{2}$ - Initial and final masses of the air within the tank, in kilograms.

$T_{1}, T_{2}$ - Initial and final temperatures of the air inflow, in Kelvin.

By (1) and key information from statement we have the following expression for the air introduced in the tank:

$m_{2}-m_{1} = \frac{V\cdot M}{R_{u}\cdot T}\cdot (P_{2}-P_{1})$ (3)

Where $P_{1}$ and $P_{2}$ are the initial and final pressures of the air inside the tank, in kilopascals.

If we know that $P_{1} = 101.33\,kPa$ , $P_{2} = 1500\,kPa$ , $R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K}$ , $T = 298.15\,K$ , $T_{1} = 318.15\,K$ , $T_{2} = 298.15\,K$ , $V = 0.1\,m^{3}$ and $M = 28.97\,\frac{kg}{kmol}$ , then the heat losses are:

$m_{2} - m_{1} = \left[\frac{(0.1\,m^{3})\cdot \left(28.97\,\frac{kg}{kmol} \right)}{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right)\cdot (298.15\,K)} \right]\cdot (1500\,kPa-101.325\,kPa)$

$m_{2}-m_{1} = 1.635\,kg$

$Q = \frac{5}{2}\cdot \left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right) \cdot \left(\frac{1}{28.97} \,\frac{kmol}{kg} \right) \cdot (1.635\,kg)\cdot (318.15\,K-298.15\,K)$

$Q = 23.461\,kJ$

A heat amount of 23.461 kilojoules is lost from the tank.

To learn more on ideal gases, we kindly invite to check this verified question: link

Answer from: AeelynRamos

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

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