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