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svet-max [94.6K]
2 years ago
10

A rigid tank contains 2 kg of N2 and 4 kg of Co2 at temperature of 25 C and 1 MPa. Find the partial pressure of each gas respect

ively?
Engineering
1 answer:
lions [1.4K]2 years ago
4 0

Answer: Partial pressures are 0.6 MPa for nitrogen gas and 0.4 MPa for carbon dioxide.

Explanation: <u>Dalton's</u> <u>Law</u> <u>of</u> <u>Partial</u> <u>Pressure</u> states when there is a mixture of gases the total pressure is the sum of the pressure of each individual gas:

P_{total} = P_{1}+P_{2}+...

The proportion of each individual gas in the total pressure is expressed in terms of <u>mole</u> <u>fraction</u>:

X_{i} = moles of a gas / total number moles of gas

The rigid tank has total pressure of 1MPa.

  • Nitrogen gas:

molar mass = 14g/mol

mass in the tank = 2000g

number of moles in the tank: n=\frac{2000}{14} = 142.85mols

  • Carbon Dioxide:

molar mass = 44g/mol

mass in the tank = 4000g

number of moles in the tank: n=\frac{4000}{44} = 90.91mols

Total number of moles: 142.85 + 90.91 = 233.76 mols

To calculate partial pressure:

P_{i}=P_{total}.X_{i}

For Nitrogen gas:

P_{N_{2}}=1.\frac{142.85}{233.76}

P_{N_{2}} = 0.6

For Carbon Dioxide:

P_{total}=P_{N_{2}}+P_{CO_{2}}

P_{CO_{2}} = P_{total}-P_{N_{2}}

P_{CO_{2}}=1-0.6

P_{CO_{2}}= 0.4

Partial pressures for N₂ and CO₂ in a rigid tank are 0.6MPa and 0.4MPa, respectively.

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klasskru [66]

Answer:

True

Explanation:

When trying to solve a frame problem in Engineering or Physics, it will typically be necessary to draw more than one body diagram.

When we have several parts of the frame or a set of frames, we have the anchor point, as well as the intersections of frames. Besides that, usually, there is a particle or rigid body together with the frame system. In this sense, usually, it is required to analyze a body diagram for the particle or rigid body suspended, as well as the intersections of the frames. So, usually, it will be required a minimum of two body diagrams.

If the system is more complex, or there are many intersections points, it will be required more than two body diagrams.

Finally, indeed, it will typically be necessary to draw many-body diagrams. 

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3 years ago
Add the following vector given in rectangular form and illustrated the process graphically A = 16+j12, B= 6+j10.4
MariettaO [177]

Answer:

A=16+j12…'B=6+j10.4

Explanation:

add the following vector given in

3 0
1 year ago
What is the resistance of a resistor if the current flowing through it is 3mA and the voltage across it is 5.3V?
Flura [38]

Answer: 1766.667 Ω = 1.767kΩ

Explanation:

V=iR

where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).

3mA = 0.003 A

Rearranging the equation, we get

R=V/i

Now we are solving for resistance. Plug in 0.003 A and 5.3 V.

R = 5.3 / 0.003

= 1766.6667 Ω

= 1.7666667 kΩ

The 6s are repeating so round off to whichever value you need for exactness.

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1 year ago
Technician A states that a brake lathe is used to make a used brake rotor surface "like new". Technician B states that a brake l
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Both Technician A and B are correct.

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2 years ago
Which sentence is correct about the exergy of an empty (pressure around zero Pascal) tank with a volume of V, located in an envi
sineoko [7]

Answer:

The correct option is;

c.  the exergy of the tank can be anything between zero to P₀·V

Explanation:

The given parameters are;

The volume of the tank = V

The pressure in the tank = 0 Pascal

The pressure of the surrounding = P₀

The temperature of the surrounding = T₀

Exergy is a measure of the amount of a given energy which a system posses that is extractable to provide useful work. It is possible work that brings about equilibrium. It is the potential the system has to bring about change

The exergy balance equation  is given as follows;

X_2 - X_1 = \int\limits^2_1 {} \, \delta Q \left (1 - \dfrac{T_0}{T} \right ) - [W - P_0 \cdot (V_2 - V_1)]- X_{destroyed}

Where;

X₂ - X₁ is the difference between the two exergies

Therefore, the exergy of the system with regards to the environment is the work received from the environment which at is equal to done on the system by the surrounding which by equilibrium for an empty tank with 0 pressure is equal to the product of the pressure of the surrounding and the volume of the empty tank or P₀ × V less the work, exergy destroyed, while taking into consideration the change in heat of the system

Therefore, the exergy of the tank can be anything between zero to P₀·V.

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