(III) A 1.00 -mol sample of an ideal diatomic gas at a...

(III) A 1.00 -mol sample of an ideal diatomic gas at a pressure of 1.00 atm and temperature of 420 $\mathrm{K}$ undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are 720 $\mathrm{K}$ and 1.60 atm. Determine $(a)$ the change in internal energy, (b) the work done by the gas, and $(c)$ the heat added to the gas. (Assume five active degrees of freedom.)

(III) A 1.00 -mol sample of an ideal diatomic gas at a
pressure of 1.00 atm and temperature of 420 $\mathrm{K}$ undergoes a
process in which its pressure increases linearly with
temperature. The final temperature and pressure are 720 $\mathrm{K}$
and 1.60 atm. Determine $(a)$ the change in internal energy,
(b) the work done by the gas, and $(c)$ the heat added to the
gas. (Assume five active degrees of freedom.)

Key Concepts

Heat Transfer

Heat transfer in a thermodynamic process refers to the energy exchanged between a system and its surroundings due to a temperature difference. In the context of the first law of thermodynamics, the heat added to the system accounts for the energy change that is not used to perform work or increase the internal energy, and is calculated based on the process path and the change in state variables.

Work in Thermodynamic Processes

Work in a thermodynamic context is the energy transferred by the system due to volume change against an external pressure. In processes where pressure is not constant, the work is calculated by integrating the pressure over the change in volume. For ideal gases, this often involves combining the ideal gas law with the specifics of the process path.

First Law of Thermodynamics

The first law of thermodynamics is a statement of energy conservation in thermodynamic processes. It relates the change in internal energy to the heat added to the system and the work done by the system, expressed as ?U = Q - W. This principle is used to analyze energy exchanges in processes involving gases.

Degrees of Freedom

Degrees of freedom in a gas refer to the independent ways in which the molecules can store energy, such as translational, rotational, and vibrational motions. In a diatomic gas at moderate temperatures, typically five degrees of freedom (three translational and two rotational) are active, which affects its internal energy and specific heat capacity.

Internal Energy of an Ideal Gas

For an ideal gas, the internal energy depends solely on temperature and the number of degrees of freedom of the gas molecules. In thermodynamic processes, the change in internal energy is determined by the change in temperature, with a specific heat capacity related to the number of active degrees of freedom.

Ideal Gas Law

The ideal gas law relates the pressure, volume, and temperature of an ideal gas through the equation PV = nRT. This law is fundamental in thermodynamics as it provides a simple model for the behavior of gases under different conditions, allowing one to determine changes in one state variable when others are varied.

Transcript

00:01 So we have a volume of the atomic gas which starts off at one atmosphere pressure and 420 kelvin temperature and expands to 1 point and goes to 1 .6 atmospheres pressure and 720 kelvin temperature.

00:23 But this happens where pressure changes linearly with temperature.

00:29 So let's say pressure goes as some p .0 plus alpha t.

00:35 Now for part a we want to find out the change in internal energy.

00:41 We know that internal energy change in internal energy is ncv d t so in this case n is 1 mole.

00:50 Cv because it's a diatomic molecule is 5 by 2 r and r is 8 .314 jolt per more kelvin and we know the delta t is 420 minus 420 which is 300 kelvin.

01:09 Giving us 6 ,240 joules.

01:13 Now for part b, we want to find out the work done.

01:16 We know that work done is integral pdv.

01:19 So we need to find volume in terms of pressure to find the work.

01:24 So before we do that, we'll have to use this information.

01:29 So we know that p is equal to p .0 alpha t.

01:31 So by substituting these two data points, we can write 1 .6 atmospheres is equal to p .0 plus alpha 720 and 1 atmosphere is equal to p .0 plus alpha into 420.

01:51 By simply subtracting these two equations we can find the value of alpha and it turns out to be 2 into 10 power minus 3 atmosphere for kelvin and then we can substitute the value of alpha back to find p .0 which turns out to be minus 0 .16 atmospheres.

02:20 So pressure is equal to 2 into 10 bar minus 3 t minus 0 .16 atmosphere, where t is in kelvin...

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