A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at 1.00 ×10^5 Pa a

step 1 A cylinder contains 0 .1 moles of an ideal gas, monotomic gas. Initially, the gas is at 1, 10 of

A cylinder contains 0.100 mol of an ideal monatomic gas....

Key Concepts

Adiabatic Process

An adiabatic process is a thermodynamic change in which the gas does not exchange heat with its surroundings. In such processes, any change in the internal energy of the gas is solely due to work done on or by the system, and the relationship between pressure and volume is defined by specific heat ratios.

Isobaric Process

An isobaric process is one in which the pressure remains constant while the volume and temperature of the gas change. This constant pressure condition is essential in many practical applications and allows for direct relationships between volume and temperature, as outlined by Charles's law.

Ideal Gas Law

The ideal gas law (PV = nRT) is a fundamental equation in thermodynamics that relates the pressure, volume, temperature, and moles of an ideal gas. It serves as the basis for analyzing gas behavior under various conditions by allowing the calculation of any one variable when the others are known.

Thermodynamic Processes

Thermodynamic processes describe the ways in which a gas can change its state, specifically how pressure, volume, and temperature interact. These include processes where one or more of these properties remain constant or follow a specific relation, forming the core of energy transfer studies in gases.

Isothermal Process

An isothermal process is characterized by a constant temperature throughout the change in state of the gas. According to the ideal gas law, if the temperature remains constant, the product of pressure and volume remains constant, which is pivotal in processes where heat exchange occurs to maintain thermal equilibrium.

Transcript

00:01 A cylinder contains 0 .1 moles of an ideal gas, monotomic gas.

00:05 Initially, the gas is at 1, 10 of the 5 pascal, occupies a volume of 2 .5 to 10 of the minus 3 meter cubed.

00:11 Find the initial temperature of the gas.

00:17 If the gas is allowed to expand to twice the volume, find a final temperature.

00:22 And if the gas is allowed to expand it twice, it's an initial volume, find the temperature in kelvin's and pressure the gas if the expansion is isothermal, isylbaric, or adiabatic.

00:31 Okay.

00:32 Let's write down what's given to us.

00:34 And start from there.

00:36 So you have the number of moles, 0 .1 moles, and you have the initial pressure, not 2, but 1 times 10 to the 5th pascal's, rdnsi, and i would need to convert.

01:03 V initial is 2 .5 times 10.

01:08 To the minus 3 meters cubed.

01:13 And our goal first a is to get the initial temperature.

01:16 This is an application of the ideal gas law.

01:19 Pv equals nrt.

01:21 If you rearrange that, then you get t is pv over nr.

01:27 And so that's 1 times 10 to the 5 times 2 .5 times 10 to the minus 3 meter cubed pascal's.

01:39 Divided by n, which is 0 .1, and then r8 .3 .1.

01:46 And if you plug that into a calculator, i'm going to double check that what i did was right.

01:57 One second.

02:03 If you plug that into a calculator and i just checked my calculations, you get 301 kelvin.

02:11 So now part b, i'm going to section this off because it might get a little crowded.

02:17 So i'm not going to write out the full calculation.

02:19 I'm just going to say things out loud.

02:21 So they're asking you to consider that your v final equals twice v initial under certain condition.

02:31 So the first condition is isothermal.

02:35 So that means p, so you have the ideal gas law.

02:41 So if t stays the same, let's ask what else stays the same? so t stays the same, r stays the same because that's a constant.

02:49 N is presumed to stay the same.

02:52 So that means pv has to say the same.

02:54 So then you can say p initial v initial equals p final, t final.

03:01 So in general, you can kind of just take whatever stays the same, put it on one side of the equation, and what changes to put it on the other side of the equation, and then say that that side is constant.

03:12 And it looks like i wrote, hold on, i think i wrote t final instead of v final.

03:18 Let me correct that.

03:27 Okay, p -final v final.

03:32 Great.

03:32 And so you can see that if the volume final goes up by a factor of two, then the pf has to go down by a factor of two to keep this product the same.

03:43 So then you get p -final is equal to half of the initial, which is 0 .5 times 10 to the 5 -pascals.

03:53 And then i'll go ahead and label this one and then the next part two and then i'll do three over here...

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