Entropy Change of Ideal Gases | Thermodynamics | (Solved Examples)

[@xo571]

didn’t even watch this video but i always appreciate the work u put into ur videos. hope all is well in ur life 😁👍

[@QuestionSolutions]

Thank you very much! I appreciate the comment. All is well, I hope yours is good too! 😊

[@milannikic3837]

thermo exam coming up, please more!!!!

[@QuestionSolutions]

I hope you do amazingly on your thermo exam!

[@iamalihussein]

Your videos are perfect ❤

[@QuestionSolutions]

Thank you! ❤

[@bhbvh4904]

Just in time for my quiz!

[@QuestionSolutions]

Best wishes on your quiz! You got this.

[@nimipen]

Hello! I have a question for the last problem you solved why was the Wout a negative?

[@QuestionSolutions]

I am not exactly sure where you're referring to, is it 11:25? If yes, we have energy in minus the energy out, and we are told there is 600kJ of work output, so the negative sign in front means we multiply the 600 by the negative sign, giving us: -(600 kJ) = -600kJ.

[@nimipen]

@@QuestionSolutions Yes it is that! Thank you so much I just got confused about the sign convention for when work is positive or negative!

[@godoffootball-24.7]

please a question concerning the second trial question, please why did you use 450k in determining the K ,since the first temperature was 300.15k

[@QuestionSolutions]

So you need an average temperature to use the equation, right? So you're just using an educated guess to figure out what the average temperature would be. You can pick 400, 450, 500K, etc, but it needs to be around there since the temperature will rise during the compression. You will get similar values regardless of the temperature picked.

[@MMMMMMMMMMBALLS]

perfect timing

[@QuestionSolutions]

Awesome!

[@kylecatman7738]

In the last example, If you said it produced 200,000 kJ we would've gotten a positive delta S also. But is that realistic?? Just because delta S is positive??

[@QuestionSolutions]

It didn't say, 200000 KJ?