n=0.081 kmol of He gas initially at 27^∘ C and pressure =2 ×10^5 N / m^2 is taken over the path

step 1 This is chapter 15 problem number 80. We're given this PV diagram for a helium gas of 3 .25 mole

N=0.081 kmol of He gas initially at 27^∘ C and pressure =2...

Key Concepts

Entropy Change in Thermodynamic Processes

Entropy is a measure of the disorder or randomness in a system. The change in entropy during a thermodynamic process can be determined by integrating the infinitesimal changes in heat transfer over the temperature. In reversible adiabatic processes, the entropy change is zero; however, in other processes, calculating the entropy change is essential for assessing the irreversibility and the overall feasibility of the state transformation.

Adiabatic Process

An adiabatic process is one in which no heat is exchanged with the surroundings (Q = 0). For an ideal gas undergoing an adiabatic process, the pressure and volume are related by pV^? = constant, where ? (gamma) is the ratio of Cp to Cv. This relationship allows the determination of the final state variables when a gas undergoes an adiabatic transformation.

Heat Capacities (Constant Pressure and Constant Volume)

Heat capacities at constant pressure (Cp) and constant volume (Cv) indicate the amount of heat required to change the temperature of a gas by one degree under constant pressure or volume, respectively. They are important parameters for determining the change in internal energy and heat transfer during thermodynamic processes, since for an ideal gas ?U depends solely on temperature and Cv.

First Law of Thermodynamics

The first law of thermodynamics states that energy is conserved and relates the change in internal energy (?U) of a system to the heat added (Q) to the system and the work done (W) by the system: ?U = Q - W. This principle is crucial for analyzing energy changes during thermodynamic processes, whether the process is isobaric, adiabatic, or otherwise.

Isobaric Process

An isobaric process is one that occurs at constant pressure. In such a process, the work done by or on the system is directly related to the change in volume, as given by the equation W = P?V. Additionally, the heat transfer in an isobaric process can be calculated using the appropriate heat capacity at constant pressure.

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, provides a fundamental relationship between pressure (P), volume (V), temperature (T), and the amount of gas (n) with the gas constant (R). It is essential in analyzing the behavior of gases under various thermodynamic processes and is applicable when the gas can be assumed to behave ideally.

Transcript

00:01 This is chapter 15 problem number 80.

00:03 We're given this pv diagram for a helium gas of 3 .25 moles.

00:11 And we're also given the fact that the process from c to a is an isothermal.

00:17 From c to a, we have an isothermal process.

00:24 Okay.

00:26 What this means is that the temperature is kept constant throughout this process.

00:31 Through this curve, which means to temperature at c equals to temperature at a and anywhere else along this path.

00:41 Right.

00:42 Now, part a, we're asked what the pressure is at point a.

00:47 Well, in order to figure that out, let's write the ideal gas law, right? so the ideal gas law for point a is going to be the pressure at that point times the volume at that point equals nr, t, at that point if you want to call it ta.

01:05 Now if you want to solve this equation, we don't, for pressure, we don't know what the temperature is.

01:13 So i'm going to write down pretty much the same equation for the point c.

01:18 So the pressure at point c times the volume at point c is going to be equal actually to the same thing, right? because the temperatures a and c are equal to each other if i write the ideal gas software the right hand side and rtc.

01:32 And since t -c -t -a -e, they're equal to each other, then we can actually drop these terms since we're not interested, then we can equate p -a -v -a to p -c -v -c.

01:45 And in order to get to p -a, we divide both sides by the volume at point a.

01:53 Then pressure at a is going to be equal to pressure at c times v -c over v -a.

02:01 That would be equal to, let's look at the pressure at c.

02:03 Pressure at c is 2 times 10 to power 2 times 10 to 5 pascoles.

02:11 And the c here is 0 .04 meter cubed divided by va would be 0 .01 meter cubed.

02:22 Now then this is 4, 8 times 10 to power 5 pascoles.

02:28 So it's quadruple, right? now, in part b, we are asked to calculate the temperature at a, temperature at b, and temperature at c.

02:42 To begin with, if we calculate the temperature at a, you know that it's going to be equal to c, right? so, temperature, these two guys are equal to each other.

02:54 Let's remember that t .a equals tc already.

02:57 Now, in order to get to t .a, for instance, we can write down the idle gas.

03:02 Law again and rta since now we know the pressure that we calculated it in order to get the t a we can divide both sides by n times r so from here temperature at a is going to be equal to pressure at a volume at a divided by n times r pressure we found it to be eight times center four or five pastcles the volume at a was point oh one meter two we're given the number of moles is 3 .25 times r being 8 .315 joules per mole kelvin.

03:38 From here, the temperature at a is found to be 296k.

03:43 All right, which means we are also found the temperature at c, right? so that is 296 kelvin.

03:52 However, we still need to figure out what temperature at b is.

03:56 So let's write down the ideal gas law for that.

03:59 Rewrite this equation for point b, so tb is going to be equal to pb times b over n times r.

04:07 Denominator is basically constant.

04:09 Nothing changes there, since the number of moles is constant.

04:14 So then let's read the pressure at b.

04:18 The pressure at b is whatever the pressure at a, right? this is an isobaric portion from a to b.

04:24 We had found this to be quadrupled.

04:27 Remember, a times 10th, chlorophy, here.

04:29 Now, then we can use that value, 8 times 10 to 4 .05 pascoles.