A cylinder contains n mol of an ideal gas that undergoes an...

A cylinder contains $n$ mol of an ideal gas that undergoes an adiabatic process. (a) Starting with the expression $W=\int P d V$ and using the expression $P V^{\gamma}=$ constant, show that the work done is $$ W=\left(\frac{1}{\gamma-1}\right)\left(P_{i} V_{i}-P_{f} V_{f}\right) $$ (b) Starting with the first law equation in differential form, prove that the work done also is equal to $n C_{V}\left(T_{i}-T_{f}\right)$. Show that this result is consistent with the equation given in part (a).

A cylinder contains $n$ mol of an ideal gas that undergoes an adiabatic process. (a) Starting with the expression $W=\int P d V$ and using the expression $P V^{\gamma}=$ constant, show that the work done is
$$
W=\left(\frac{1}{\gamma-1}\right)\left(P_{i} V_{i}-P_{f} V_{f}\right)
$$
(b) Starting with the first law equation in differential form, prove that the work done also is equal to $n C_{V}\left(T_{i}-T_{f}\right)$. Show that this result is consistent with the equation given in part (a).

Key Concepts

Adiabatic Process

An adiabatic process is one in which no heat is exchanged with the surroundings, meaning that all changes in the internal energy of the system are due solely to work done. For an ideal gas undergoing an adiabatic process, this condition leads to specific relationships between pressure, volume, and temperature, notably P V^? = constant.

Thermodynamic Work

Thermodynamic work in reversible processes is computed as the integral W = ? P dV. In the context of an adiabatic process, the pressure P can be expressed as a function of volume using the relation P V^? = constant, which allows the integral to be evaluated to give an expression for the work done in terms of the initial and final states.

Adiabatic Equation (P V^? = constant)

This equation embodies the constraint on an ideal gas undergoing an adiabatic process. It relates pressure and volume with the adiabatic index ? and ensures that as the gas expands or compresses adiabatically, changes in pressure and volume adhere to this fixed product. This relation is key to transforming the work integral into an expression involving state variables.

First Law of Thermodynamics

The first law of thermodynamics, stated in differential form as dU = dQ - dW, governs energy conservation in thermodynamic systems. For an adiabatic process, dQ = 0, thus any change in the internal energy of the system is entirely due to the work done on or by the system, linking the change in internal energy directly to W.

Specific Heat at Constant Volume (C_V)

C_V is the amount of heat required to raise the temperature of one mole of a gas by one degree at constant volume. In an adiabatic process where no heat is exchanged, the change in internal energy (n C_V ?T) directly reflects the work done on or by the gas. This provides an alternative expression for the work done, demonstrating consistency with the result obtained from integrating the pressure-volume relation.

Transcript

00:01 In this question, we are asked to derive the work done on an adiabetic process.

00:07 I do guess that undergoes an adiabetic process.

00:11 So in part a, okay, so we are given adibetic process.

00:32 The equation is pv to the gamma equals to constant, and then starting, and then this constant, i'm just going to call it k, and then w, the work down by guess, is given to be pdv.

00:52 So p is k over v gamma db, and then you integrate from v initial to v final.

01:05 So you have integral from the i to vf, k times v to the negative gamma, db, k, then you have k times one minus gamma times v to the power 1 minus gamma, and then from v i to vf, okay, then you can put in the limits.

01:46 So i'll flip the gamma, by doing so you flip the things inside, and then we have to pv to the gamma, right? so this means that you can also be pi, vi to the power of gamma.

02:12 It can also be equal to p -final, v -final to the power of gamma, okay? so this is what we are going to do.

02:20 So when you see vi, you put the p -i -v -i version.

02:31 Then you see the vf, you put the vf version.

02:39 Okay, then you group the v -i and the vf together, and you can see the v -f version.

02:43 That the v -i -a -to -the -gama and v -i -to -the -negative gamma, they can out, right? so we have one over, gamma minus -1, p -i -v -i -v -f, okay? right, so this is how we arrive at this expression for the work done by guess in an diabetic process.

03:11 Okay, then in part b, we are asked to start from the, using the first law and show that the result is consistent with part a.

03:19 So from first law, thermodynamics, okay, we have d .e.

03:37 Internal is equal to dq minus d .w...

Please give Ace some feedback