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kap26 [50]
2 years ago
8

What is the change in entropy of 250 g of steam at 100 ∘c when it is condensed to water at 100 ∘c?

Chemistry
1 answer:
Veronika [31]2 years ago
4 0

Entropy change is the randomness of the thermodynamic system. The change in the entropy when the water is condensed is 1573 J/K.

<h3>What is the entropy change?</h3>

An entropy change is a product of mass and the ratio of the enthalpy of vaporization to the temperature.

Given,

Enthalpy of vapourization =  2257 J/gK

The temperature in kelvin = 373 k

The change in entropy is calculated as:

\begin{aligned} \rm \Delta S &= \rm m (\dfrac{\Delta H_{vap}}{T})\\\\&= 260 (\dfrac{2257}{373})\\\\&= 1573.24\;\rm J/K\end{aligned}

Therefore, the change in entropy is 1574 J/K.

Learn more about entropy here:

brainly.com/question/14798359

#SPJ4

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A mixture of two gases with a total pressure of 2.12 atm contains 0.70 atm of Gas A. What is the partial pressure of Gas B?
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Answer:

1.42

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6 0
3 years ago
One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
3 years ago
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Explanation:

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