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Tomtit [17]
3 years ago
7

A system of ideal gas at 22°C undergoes an ischoric process with an internal energy decrease of 4.30 × 10 3 4.30×103 J to a fina

l temperature of 16°C. What is the approximate change in entropy?[Hint: use the average temperature]
Physics
1 answer:
Komok [63]3 years ago
5 0

Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

Given that,

Temperature = 22°C

Internal energy U=4.30\times10^{3}\ J

Final temperature = 16°C

We need to calculate the approximate change in entropy

Using formula of the entropy

\Delta S=\dfrac{\Delta U}{T}

Where, \Delta U = internal energy

T = average temperature

Put the value in to the formula

\Delta S=\dfrac{-4.30\times10^{3}}{\dfrac{22+273+16+273}{2}}

\Delta S=-14.72\ J/K

Hence, The approximate change in entropy is -14.72 J/K.

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The ray diagram for the given object consists of 2 cm height of object, 4 cm object distance and 3 cm focal length.

<h3>Image formed by a diverging lens</h3>

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<h2>MARK BRAINLIEST</h2>

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