What is the change in entropy (∆S) when 1.32 g of propane (C3H8) at 0.100 atm pressure is compressed by a factor of five at a constant temperature at 20°C? Assume that propane behaves as an ideal gas.
(a) ∆S = +13 J/K
(b) ∆S = -13 J/K
(c) ∆S = - 0.40 J/K
(d) ∆S = + 0.40 J/K

Question

What is the change in entropy (∆S) when 1.32 g of propane (C3H8) at 0.100 atm pressure is compressed by a factor of five at a constant temperature at 20°C? Assume that propane behaves as an ideal gas. 
(a) ∆S = +13 J/K
(b) ∆S = -13 J/K 
(c) ∆S = - 0.40 J/K
(d) ∆S = + 0.40 J/K

Accepted Answer

Hello. Everyone in this video we're being told that at a constant temperature of 25 degrees. We want to go ahead and calculate for the delta s. Or the change in entropy when 5.75 g of butane so has formula of C four H 10 At a given pressure of .250 80M expand by a factor of four. We will also consider that butane behaves at an ideal gas. So to calculate the change of entropy for the expansion at a constant temperature, given the number of moles of an ideal gas. Go ahead and use the formula for delta S. Equaling to the number of moles times our deal gas law constant times the natural log of V two over V one. So to remind ourselves V two is our final volume and V one is our initial volume. So first what we can do is utilize the given mass of beauty into 5.75 g, convert that into the number of moles of routine. So we have the molecular weight of butane equaling 258.1-2 units being g per mole. And this for butane. So this value can either be solved on your own by using the P. R. Table or we can go ahead and also find this online or is even given to you by a professor. This is just the value that I have and we're gonna go ahead and use this for our calculations. So we're gonna go ahead and just say that we're solving for n. Which is going to be simplified for the number of moles of routine. Again, we're starting off with the mass that's given. So that's 5. grounds of C four H 10. Again, that's just the molecular formula for beauty. We can do some dimension analysis to convert our grams into moles. So we're gonna go ahead and use the molecular weight. So for every one mole of butane we have 58.1 to 2 g. You can see here that the grams unit will cancel leaving us with just the moles of beauty. So once I put everything into my calculator I see that I get a value of 9.8930 times 10 to the negative two moles of c four H 10. Alright. And since boutin gas expands by a factor of four, we know that the final volume is going to be. So V two is for a final volume is equal to four times our initial volume. So basically in our equation we can substitute in our V two for four V one. So now we go ahead and solve for the change of entropy using our formula then so we have again, delta S. Is equal to the number of moles. Multiply bar gas law constant, multiplied by the natural log of we said V two over V one. We already have this value here. So let's go ahead. Put that in the delta S. Is equal to n our natural log of Well now is V. Or four times V. One over V. One. You can see here in this equation that V1 will actually cancel each other out, leaving us with just natural log of four. Let's go ahead and plug in some values then. So we have the number of moles which we software and green. That is 9.89 right out the whole value here. So again, 9.8930 times 10 to the - moles. And then our gas law constant R. That is 8.314 units being Joel, jules per mole times kelvin. And then we have the natural log of four. So once I put everything into my calculator, We see that the numerical value that I get out is running to 1.14. So we have three significant figures. We'll see her on the line above. So this right here that the moles will go ahead and cancel. So for my units then that she's going to be simply jewels her kelvin. So I have all my answer choices from A to D. We can see that we have this value that we solve for being as a choice. A. So answer choice is going to be my final answer for this problem

What is the change in entropy (∆S) when 1.32 g of propane (C3H8) at 0.100 atm pressure is compressed by a factor of five at a constant temperature at 20°C? Assume that propane behaves as an ideal gas. (a) ∆S = +13 J/K (b) ∆S = -13 J/K (c) ∆S = - 0.40 J/K (d) ∆S = + 0.40 J/K

Verified Solution

Key Concepts

Entropy (S)

Ideal Gas Law

Change in Entropy for Compression