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weqwewe [10]
2 years ago
11

Chegg an insulated rigid tank contains 12 kg of air at 200 kpa and 500 c. The tank valve is opened, and pressure reduces to 75 k

pa through an isentropic process. Determine the final temperature (k)
Chemistry
1 answer:
alekssr [168]2 years ago
8 0

In thermodynamics, the adiabatic and reversible system is the isotropic process. The final temperature of the system is 311.08 degrees celsius.

<h3>What are isotropic conditions?</h3>

Isotropic conditions are the process of the thermodynamic system that has no net transfer of heat.

Given,

Initial temperature = 773.15 Kelvin

Initial pressure = 200

Final pressure = 75

The final temperature is calculated as:

\rm \dfrac{T_{2}}{T_{1}} = \rm \dfrac{P_{2}}{P_{1}} ^{y- \frac{1}{y}}

Substituting values:

\begin{aligned}\rm \dfrac{T_{2}}{773} &= \dfrac{75}{200}^{1.4- \frac{1}{1.4}}\\\\&= 584.08 \;\rm kelvin\end{aligned}

Therefore, 584.08 K or 311.08 degrees celsius is the final temperature.

Learn more about the isotropic system here:

brainly.com/question/13497738

#SPJ4

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Consider the equation ∆G = –nFE, where F = 9.648533289 x 104 C/mol. Given that n is the mole of e– transferred and the following
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How much heat is required to vaporize 31.5 gg of acetone (C3H6O)(C3H6O) at 25 ∘C∘C? The heat of vaporization for acetone at this
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Answer:

≅ 16.81 kJ

Explanation:

Given that;

mass of acetone = 31.5 g

molar mass of acetone = 58.08 g/mol

heat of vaporization for acetone = 31.0 kJ/molkJ/mol.

Number of moles = \frac{mass}{molar mass}

Number of moles of acetone = \frac{31.5}{58.08}

Number of moles  of acetone = 0.5424 mole

The heat required to vaporize 31.5 g of acetone can be determined by multiplying the number of moles of acetone with the heat of vaporization of acetone;

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The heat required to vaporize 31.5 g of acetone = 0.5424 mole × 31.0 kJ/mol

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3 years ago
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