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ratelena [41]
3 years ago
10

An insulated 8m* rigid tank contains air, modeled as an ideal gas, at 600kPa and 400K. A valve connected to the tank is opened a

nd air is allowed to escape until the pressure drops to 200kPa. The air temperature is held constant during the process by an electric heater placed in the tank. Determine the electrical energy supplied to the air during this procesS. Internal energy of the initial air is U = 286.1 6hĺ kg ; The enthalpy of the exiting air is Hvit400.98HJ/ kg
Chemistry
1 answer:
garik1379 [7]3 years ago
8 0

Explanation:

We assumes that all the electrical energy produced will be converted into the heat energy.

    Electrical energy input = Heat energy absorbed by object = Heat energy required = Q

As,        Q = U - W

                = U - RT ln \frac{P_{2}}{P_{1}}

Since, it is given that U = 286.1 kJ/kg, P_{1} is 600 kPa, P_{2} is 200 kPa, and T is 400 K.

Therefore, putting these given values into the above formula is as follows.

          Q = U - RT ln \frac{P_{2}}{P_{1}}

              = 286.1 kJ/kg - 8.314 kJ/kmol K \times 400 K \times ln \frac{200}{600}

              = 286.1 kJ/kg - \frac{8.314}{28.97} kJ/kg K \times 400 K \times ln \frac{200}{600}

              = 412.27 kJ/kg

Thus, we can conclude that the electrical energy supplied to air is 412.27 kJ/kg.

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en un recipiente se tiene 800 g de una solución al 35% en masa de ácido sulfuroso, de la cual se evapora 80ml de agua. ¿cuál es
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The mass percent of sulfurous acid in the new solution : 38.9%

<h3>Further explanation</h3>

<em>In a container you have 800 g of a 35% by mass solution of sulfurous acid, from which 80 ml of water evaporates. What is the mass percent of sulfurous acid in the new solution? data: density of water is 1g / ml.</em>

<em />

solution 1

composition :

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\tt 0.35\times 800~g=280~g

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solution 2(new solution)

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\tt \dfrac{280}{720}\times 100\%=38.9\%

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