Entropy-Isobaric Expansion of a gas to thermal equilibrium

[TheTourist]

One mole of an ideal gas undergoes an isobaric expansion to reach thermal equilibrium with a surrounding heat bath. In doing so, it expands by a factor of two in volume. The internal energy of the gas is described by, U=3/2RT.
Calculate the resulting change in entropy of the gas, the heat bath, and the Universe.

Homework Equations

$$\delta$$S=$$\frac{delta Q}{T}$$

I understand all the theory behind this, I just can't seem to derive an equation which would allow me to obtain an answer with the information given, I always seem the need a temperature. I have tried using the above equations for U and delta S, and using the 1st of law of thermodynamics, but can't seem to get anywhere.

Thanks

 

[Andrew Mason]

One mole of an ideal gas undergoes an isobaric expansion to reach thermal equilibrium with a surrounding heat bath. In doing so, it expands by a factor of two in volume. The internal energy of the gas is described by, U=3/2RT.
Calculate the resulting change in entropy of the gas, the heat bath, and the Universe.

Homework Equations

$$\delta$$S=$$\frac{delta Q}{T}$$

I understand all the theory behind this, I just can't seem to derive an equation which would allow me to obtain an answer with the information given, I always seem the need a temperature. I have tried using the above equations for U and delta S, and using the 1st of law of thermodynamics, but can't seem to get anywhere.

Thanks

The change in entropy is:

$$\Delta S = \int_{rev} dQ/T$$

The reversible path is one in which external pressure remains constant and heat flows reversibly from the surroundings to the gas (eg. via a Carnot heat engine/pump). dQ = C_pdT. So the change in entropy of the gas is:

$$\Delta S = \int_{rev} C_pdT/T$$

For the surroundings, which remain at constant temperature, the heat flow is equal and opposite to the heat flow into the gas:

$$\Delta S = -\Delta Q/T_{surr} = -\frac{1}{T_{surr}\int_{rev}C_pdT_{gas}$$

You just have to work out the change in temperature of the gas.

AM

 

[Deadlyhedley]

I have this exact question to do for tomorrow. I don't understand what this question is asking. Is there any more steps to the solution that what you have already provided?
Thanks

 

[Andrew Mason]

I have this exact question to do for tomorrow. I don't understand what this question is asking. Is there any more steps to the solution that what you have already provided?
Thanks

First you work out the temperature change.

Then you evaluate the integral $$\int_{T_i}^{T_f} dQ/T$$ for the gas. That gives you the change in entropy of the gas.

Then you evaluate $Q_{surr}/T_{surr} = -Q_{gas}/T_{surr}$ to determine the change in entropy of the surroundings.

Then you add them together to get the change in entropy of the universe.

AM

 

[rwisz]

for your question! Let's break down the problem and see if we can come up with a solution.

First, let's define some terms to make sure we are on the same page. Entropy (S) is a measure of the disorder or randomness in a system. Isobaric expansion means that the pressure of the gas remains constant during the expansion process. Thermal equilibrium means that the gas and the heat bath have reached the same temperature.

Now, let's consider the first law of thermodynamics, which states that the change in internal energy (delta U) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system. In this case, since the gas is expanding, work is being done by the system, so we can write:

delta U = Q - W

Since this is an isobaric expansion, the work done by the system can be calculated as:

W = P(delta V)

Where P is the constant pressure and delta V is the change in volume. In this case, delta V is given as a factor of two, so we can write:

W = P(2V - V) = PV

Now, we can substitute this into the first law of thermodynamics equation to get:

delta U = Q - PV

Next, we can use the ideal gas law to relate pressure, volume, and temperature:

PV = nRT

Where n is the number of moles of gas and R is the gas constant. We can rearrange this equation to solve for P:

P = (nRT)/V

Now, we can substitute this into our first law of thermodynamics equation to get:

delta U = Q - (nRT)

Since the gas undergoes an isobaric expansion, the pressure remains constant, so the heat added to the system is equal to the change in internal energy. Therefore, we can rewrite our equation as:

delta U = delta Q = (nRT)

Now, we can use the equation for entropy (delta S = delta Q/T) to calculate the change in entropy of the gas:

delta S = (nRT)/T = nR ln(2)

This is the change in entropy of the gas. However, we also need to consider the change in entropy of the heat bath and the universe. Since the heat bath is at thermal equilibrium with the gas, it will also experience a change in entropy equal to the change in entropy of the