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Annette [7]
1 year ago
8

an ideal otto cycle with argon as the working fluid has a compression ratio of 8. the minimum and maximum temperatures in the cy

cle are 540 and 2350 r. account for the variation of specific heats with temperature. the properties of argon are cp
Chemistry
1 answer:
Verizon [17]1 year ago
8 0

The specific heat of gases can be taken roughly as a constant for differences in the order of 100⁰ C from ambient. Variation is crucial and cannot be disregarded for temperatures of more than, let's say 500 ⁰C or 1000⁰C.

  1. The ideal gas constant is the difference between cp and cv for low pressures (the ideal gas).
  2. The classical statistical physics principles for ideal non-interacting gases are found in any university physics primer.
  3. The demonstration that pV/T = constant is typically included in texts on macroscopic thermodynamics suggests that while temperature changes depend on specific heats, the opposite is also true.
  4. Real gas behavior requires a more complex explanation.
  5. As a result, we employ two techniques to determine the specific heat of gases: at constant volume and constant pressure.
  6. The value of the heat capacity at constant pressure is always greater than the value of the heat capacity at constant volume because the former also takes into account the value of the heat energy required to expand the substance against the constant pressure as its temperature rises.

To learn about Real gas

<u>brainly.com/question/17355868</u>

#SPJ4

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To create hydrogen and oxygen from water, you would need to perform which type of reaction?
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The following compound is used as an adhesive in the oil and gas industry.
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See Explanation

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3 0
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Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose
Elanso [62]

Answer:

The leftover is 15 g of oxygen rather than liquid hexane as the fuel limits the reaction.

Explanation:

Hello!

In this case, since the described combustion reaction is:

C_6H_{14}(l)+\frac{19}{2} O_2(g)\rightarrow 6CO_2(g)+7H_2O(g)

Thus, since 5.2 g of hexane (86.2 g/mol) is reacted with 33.0 g of oxygen (32.0 g/mol) we can compute the mass of hexane that was actually consumed via stoichiometry with oxygen (1:19/2 mole ratio):

m_{C_6H_{14}}^{consumed \ by\ O_2}=33.0gO_2*\frac{1molO_2}{32.0gO_2}*\frac{1molC_6H_{14}}{\frac{19}{2}gO_2 }  *\frac{86.2gC_6H_{14}}{1molC_6H_{14}} \\\\m_{C_6H_{14}}^{consumed \ by\ O_2}=9.36gC_6H_{14}

It is proved then than the hexane won't have any leftover but oxygen does, as shown below:

m_{O_2}^{consumed \ by\ C_6H_{14}}=5.2gC_6H_{14}*\frac{1molC_6H_{14}}{86.2gC_6H_{14}} *\frac{\frac{19}{2}molO_2 }{1molC_6H_{14}} *\frac{32.0gO_2}{1molO_2} \\\\m_{O_2}^{consumed \ by\ C_6H_{14}}=18g

It means the leftover of oxygen is:

m_{O_2}^{leftover}=33g-18g\\\\m_{O_2}^{leftover}=15g

Regards!

6 0
3 years ago
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