Reversible, adiabatic expansion - Entropy Change

[laser1]

Homework Statement

Calculate the change in entropy of the universe

Relevant Equations

$\Delta S = \int_i^f{\frac{dQ}{T}}$

Answer for (d) is 0, answer for (e) is not.

Firstly, I don't get why (e) is not zero. It says "the same expansion" so that expansion is reversible. Reversible processes -> entropy = 0?
Secondly, part (e) seems to be the exact same as (d) so I'm not sure why it's different!

Thanks in advance

 

[Philip Koeck]

Homework Statement: Calculate the change in entropy of the universe
Relevant Equations: $\Delta S = \int_i^f{\frac{dQ}{T}}$

View attachment 351755
Answer for (d) is 0, answer for (e) is not.

Firstly, I don't get why (e) is not zero. It says "the same expansion" so that expansion is reversible. Reversible processes -> entropy = 0?
Secondly, part (e) seems to be the exact same as (d) so I'm not sure why it's different!

Thanks in advance

(d) is reversible, (e) is not. (e) is actually a free expansion.
In (d) the system's temperature changes, whereas in (e) it doesn't. That's another difference between them.

 

[laser1]

(d) is reversible, (e) is not. (e) is actually a free expansion.
In (d) the system's temperature changes, whereas in (e) it doesn't. That's another difference between them.

Ah okay, but the phrasing "same expansion" seems to suggest (e) is reversible, the same as (d), no?

 

[Philip Koeck]

Ah okay, but the phrasing "same expansion" seems to suggest (e) is reversible, the same as (d), no?

They should have written "expansion with the same initial and final volume".
(e) is clearly a free expansion so it can't be exactly the same as (d).
In (e) no work is done by the the gas, in (d) there is.

 

[Andrew Mason]

Homework Statement: Calculate the change in entropy of the universe
Relevant Equations: $\Delta S = \int_i^f{\frac{dQ}{T}}$

View attachment 351755
Answer for (d) is 0, answer for (e) is not.

Firstly, I don't get why (e) is not zero.

The problem with the question as provided is that it does not state that it is an ideal gas.

If it is an ideal gas, there will be no temperature change as $\Delta U=nC_v\Delta T=0$. So the change of entropy is the change of entropy over a reversible adiabatic expansion to twice the volume ($\Delta S=0$, $T=T_a$) plus the change in entropy of a reversible heat flow into the gas at constant volume to bring the temperature up to the original 0C/273K. $\Delta S_{gas}=\int_{T_a}^{273}\frac{nC_vdT}{T}=nC_v\ln\left(\frac{273}{T_a}\right)$. One just has to determine the temperature after the adiabatic expansion $T_a$ (using the adiabatic condition and $\gamma$ of the gas). Alternatively, you could use a reversible expansion at constant temperature (constant U) and equate W and Q so $\Delta S_{gas}=\int_{V_i}^{V_f}\frac{PdV}{T}=\int_{V_i}^{V_f}\frac{nRTdV}{TV}=nR\int_{V_i}^{V_f}\frac{dV}{V}=nR\ln\left(\frac{Vf}{V_i}\right)=nR\ln(2)$. Change in entropy of the surroundings is $\Delta S=-\Delta Q_{gas}/273$

But if it is a real gas in which there are inter-molecular forces, potential energy of the molecules will increase with free expansion and temperature of the expanded gas will decrease despite the lack of work being done by the gas. So the reversible path for determining change in entropy will depend on its final temperature and heat capacity.

 

[Chestermiller]

The problem with the question as provided is that it does not state that it is an ideal gas.

If it is an ideal gas, there will be no temperature change as $\Delta U=nC_v\Delta T=0$. So the change of entropy is the change of entropy over a reversible adiabatic expansion to twice the volume ($\Delta S=0$, $T=T_a$) plus the change in entropy of a reversible heat flow into the gas at constant volume to bring the temperature up to the original 0C/273K. $\Delta S_{gas}=\int_{T_a}^{273}\frac{nC_vdT}{T}=nC_v\ln\left(\frac{273}{T_a}\right)$. One just has to determine the temperature after the adiabatic expansion $T_a$ (using the adiabatic condition and $\gamma$ of the gas).

This was an interesting illustration that all reversible paths lead to the same change in entropy, even this unusual two-step process, but I feat that this example will only serve to confuse the OP>

But if it is a real gas in which there are inter-molecular forces, potential energy of the molecules will increase with free expansion and temperature of the expanded gas will decrease despite the lack of work being done by the gas. So the reversible path for determining change in entropy will depend on its final temperature and heat capacity.

The reversible path is whatever we specify iii to be. Some reversible paths only involve the heat capacity in the ideal gas limit (which we know) as a function of temperature.