Consider a twofold expansion of 1 mol of an ideal gas at 25 °C in...

Question

Consider a twofold expansion of 1 mol of an ideal gas at 25 °C in the isolated system shown in Figure 18.1. <QUESTION REFERENCES FIGURE 18.1> 
(a) What are the values of ∆H, ∆S, and ∆G for the process?

Accepted Answer

Hello everyone today. We have the following problem. The volume of a two most sample of an ideal gas was tripled at a constant temperature of 50°C, determine the values of the entropy change entropy change and our Gibbs free energy for the expansion. So what we want to first start off with is our entropy change. Our entropy changes actually going to be equal to zero and this is because entropy is a function of temperature and temperature is constant, so it's not really changing. Secondly we can solve for our entropy change and this is going to be given by the formula that we are that we have our moles of gas times of gas constant times the natural log of our final volume over our initial So what is that gonna look like? Well We have two moles of gas. R gas constant is 8. joules per kelvin mole And it's gonna be the natural log of three since the volume was tripled. So the ratio is gonna be by three. So we just take the natural log of that. We saw this into our calculator, We're going to get 18. jewels per kelvin. And that's gonna be our entropy change last but not least. We're gonna calculate the gibbs free energy and that is actually an equation that's gonna involve all these variables. So it's gonna be the change in our entropy minus temperature times the change in our entropy. And so we said that our entropy change was zero. So we're gonna do zero minus our temperature which is 50 degrees Celsius. However we need this in terms of kelvin. So we're just gonna add to 73.15 And then we're going to multiply that by our entropy change which was 18. jewels per kelvin. And when we do that we're going to get negative 5,903.21 jewels. However this needs to be in terms of kilo jewels. So what we're gonna do because we're going to take this and multiply it by the conversion factor, the jewels the conversion factor that one killer jewel is a go to tend to the third jewels. And so when our units cancel that we will be left with negative 5. killer jewels. And so we have our entropy change here are entropy change here and our gibbs free energy. And with that we have solved the problem overall, I hope this helped. And until next time

Consider a twofold expansion of 1 mol of an ideal gas at 25 °C in the isolated system shown in Figure 18.1. (a) What are the values of ∆H, ∆S, and ∆G for the process?

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