Entropy Expand 1.00 mol of an monatomic gas initially at 5.00 kPa and 600 K from initial volume

step 1 So here for part A, we need to find the final pressure. So the final pressure can be found by us

Entropy Expand 1.00 mol of an monatomic gas initially at...

Entropy Expand 1.00 mol of an monatomic gas initially at 5.00 $\mathrm{kPa}$ and 600 $\mathrm{K}$ from initial volume $V_{i}=1.00 \mathrm{m}^{3}$ to final volume $V_{f}=$ 2.00 $\mathrm{m}^{3} .$ At any instant during the expansion, the pressure $p$ and volume $V$ of the gas are related by $p=5.00 \exp \left[\left(V_{i}-V\right) / a\right],$ with $p$ in kilopascals, $V_{i}$ and $V$ in cubic meters, and $a=1.00 \mathrm{m}^{3} .$ What are the final (a) pressure and (b) temperature of the gas? (c) How much work is done by the gas during the expansion? (d) What is $\Delta S$ for the expansion? (Hint: Use two simple reversible processes to find $\Delta S.$)

Entropy
Expand 1.00 mol of an monatomic gas initially at 5.00 $\mathrm{kPa}$ and 600 $\mathrm{K}$ from initial volume $V_{i}=1.00 \mathrm{m}^{3}$ to final volume $V_{f}=$ 2.00 $\mathrm{m}^{3} .$ At any instant during the expansion, the pressure $p$ and volume $V$ of the gas are related by $p=5.00 \exp \left[\left(V_{i}-V\right) / a\right],$ with $p$ in kilopascals, $V_{i}$ and $V$ in cubic meters, and $a=1.00 \mathrm{m}^{3} .$ What are the final (a) pressure and (b) temperature of the gas? (c) How much work is done by the gas during the expansion? (d) What is $\Delta S$ for the expansion? (Hint: Use two simple reversible processes to find $\Delta S.$)

Key Concepts

Ideal Gas Law

The ideal gas law (pV = nRT) is a fundamental relationship in thermodynamics linking pressure, volume, and temperature for a given amount of an ideal gas. It is used to determine one state variable when the others are known and is critical for analyzing thermodynamic processes, as it provides the baseline connection among the gas’s macroscopic properties.

Reversible Processes

A reversible process is an idealized thermodynamic transformation that occurs infinitely slowly so that the system remains in equilibrium at every stage. This concept is important because reversible processes allow for precise calculation of work and heat transfer, and they provide a means to calculate state function changes, such as entropy, by conceptually connecting states through ideal paths.

Work in Thermodynamic Processes

In thermodynamics, the work done by or on a system during a volume change is often computed by integrating the pressure with respect to volume (W = ? p dV). When the pressure changes according to a given path relation, this integration determines the total work performed during the expansion or compression, which is crucial for energy analysis in any process.

Entropy and Its Calculation

Entropy is a state function that represents the degree of disorder or randomness in a system. It is defined for a reversible process through the integral of the heat transfer divided by temperature (?S = ? (?Q_rev/T)). Calculating entropy changes by considering reversible paths simplifies the process, as the overall change depends only on the initial and final states, irrespective of the actual path taken.

Path Dependence versus State Functions

In thermodynamics, certain properties like work and heat are path dependent, meaning that the amount transferred can vary with the process path used to transition between states. In contrast, state functions such as internal energy, enthalpy, and entropy depend solely on the initial and final states. This distinction is crucial for simplifying the evaluation of thermodynamic changes, as one can choose an easier reversible path to compute the change in a state function like entropy.

Transcript

00:01 So here for part a, we need to find the final pressure.

00:03 So the final pressure can be found by using the given relationship, 5 .00 kilopascals multiplied by e to the v initial minus v final, so essentially the change in volume, divided by a.

00:24 And so we can solve the final pressure would be equal to 5 .00 kilopascal.

00:32 I'll just use the exponential function.

00:35 The exponential function simply means that whatever is inside the parentheses will be the power with a base of e.

00:42 And this is simply to essentially just write it a little bit neater.

00:48 So e essentially to the 1 .00 meter cubed minus 2 .00 meter cubed divided by 1 .0 meter cubed divided by 1 .0.

01:03 0 .00 meter cubed and we find that the final pressure is 1 .84 kilopascals.

01:14 For part b, this would be your answer for part a, the final pressure.

01:20 For part b, we can use the ratio form of the gas law to find the final temperature of the gas.

01:26 So we can say temperature final equals temperature initial and then we'll multiply this by pressure final, volume final, divided by pressure initial, volume initial.

01:38 And at this point we can solve.

01:40 So it'll be 600 kelvin multiplied by 1 .84 kilopascals multiplied by 2 .00 meters cubed.

01:54 This would be divided by our initial pressure of 5 .00 kilopascals multiplied by 1 .00 meters cubed and we find that the final temperature is 441 kelvin...

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