Expand 1.00 mol of an monatomic gas initially at 5.00 kPa and 600 K from initial volume Vi=1.0

Expand 1.00 mol of an monatomic gas initially at 5.00 kPa...

Key Concepts

Ideal Gas Law

The Ideal Gas Law (pV = nRT) is a fundamental relationship in thermodynamics relating the pressure, volume, and temperature of an ideal gas. It is used to connect macroscopic properties in problems involving state changes of gases and provides a basis for relating any two state variables when one is changed and the others are known.

Pressure-Volume Work

In thermodynamics, work done by or on a gas during expansion or compression is calculated by integrating the pressure with respect to volume change (W = ? p dV). This concept is especially crucial in processes where pressure varies with volume, requiring the evaluation of an integral rather than a simple multiplication of pressure and volume.

Reversible Processes

A reversible process is an idealized thermodynamic process that occurs infinitely slowly so that the system remains in equilibrium at all times. This concept is important because reversibility implies maximum work output and allows for the use of state functions, like entropy, to analyze changes without needing the full path details.

Exponential Process Relationships

In some thermodynamic processes, pressure can be a specific function of volume, such as an exponential relationship. Understanding and working with these relationships is essential, as it affects the integration required for work calculation and influences the way temperature and other state variables evolve during the process.

Entropy Change of Ideal Gases

Entropy is a state function that measures the disorder or randomness of a system. For ideal gases, the change in entropy can be calculated using known end states of temperature and volume (or pressure) by employing formulas derived from the Sackur-Tetrode equation or by integrating the reversible path. This becomes particularly relevant when the process path is not a simple isothermal or adiabatic one, prompting a breakdown into simpler reversible steps.

Transcript

00:01 Now here in this problem we are given that mono atomic gas with number of molds in 2b1 is at volume 1 meter cube pressure initial pressure we have 5 kylpascal and initial temperature we have 600 kelvin.

00:14 Now this mono atomic gas is expanded and here the final volume becomes 2 meter cube and the pressure and the volume is related by the equation p is equal to 5.

00:30 Times exponential to the power vi minus v divide by a this is a relationship between the volume and the pressure now the value of a we are given to be asked one meter cube it is given in the equation now there are four parts in the equation in the part we have to calculate the pressure final pressure and in the b part we have to calculate the final temperature so let us first solve these two problems part a so if you see this equation from here this from by using this equation we can write that the final pressure will be equals to five times a to the power vi minus this volume would become final volume divide by a so it will be equals to five times a to the power now initial volume we are given as one meter cube that would be one minus final volume becomes two that would it to divide by a so a we have one so one it will be equals to five divide by e and from here we get the pressure to be 1 .84 kioskal so this is the answer for the a part so this is 1 .84 kiosk let me write here it will be 1 .84 kilo pascal now next we're going to solve the b part of the problem where we you have to calculate the final temperature of the monoatomic gas now for a mono atomic idle gas we can write that pv divide by t equals to constant and from here we can write that final pressure times the final volume divide by the final temperature will be equals to the initial pressure times the initial volume divide by the initial temperature now we have all the values we can plug it here to calculate the final temperature so let us put the values so final pressure we have 1 .84 multiplied with the final volume we have two divide by the final temperature that we have to calculate is equals to now initial pressure we are given us 5 kilo pascal multiplied with initial volume we have one divide by initial temperature we have 600 so from here we'll get the final temperature to be nearly 4401 1 kelvin so this is the answer for the b part so let me write here the final temperature becomes 4 4 1 kelvin now next we're going to solve the c part of the problem where we have to calculate the work done by the gas during the expansion process so we're going to solve the c part so we know that the work done by the gas is given by work is equals to integration pdv this is generalized expression and volume we have initial volume to final volume now pressure the expression for pressure we are already given in the equation to be five times a to the power v i minus v divide by a d and this d would remain as it is integration of initial volume we have in the equation to be as one and final volume we have two now this integration can also be written as integration 5 i can take outside and here it would become a to the power v i divide by a multiplied with a to the power negative v divide by a dv we can also write this way also and this is from limit to 2 now this is a constant so we can take this term outside the integration so five times a to the power v i divide by a now list we can take the integration of this function and it will be minus a times a to the power minus v divided by a and the limit is from 1 to 2 now we can put the limits so 5 times a to the power now we have 1 beta cube so 1 divide by 1 times this minus a i can take outside and a we have 1 so this would remain remain as negative only because this negative 1 i've taken outside now we can put the limits so a to the power minus 2 divide by this would be 1 minus a to the power minus 1 divide by 1 and on further simplification will get the answer to be 3 .16 kilojoule so this is the work done by the gas during the expansion process and the answer for the c part of the problem next we're going to solve the d part where we have to calculate the entropy change during the expansion process...

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