1.

Give an expression for work done in an isothermal process.

Answer» <html><body><p></p>Solution :Isothermal process: It is a process in which the <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> <a href="https://interviewquestions.tuteehub.com/tag/remains-621920" style="font-weight:bold;" target="_blank" title="Click to know more about REMAINS">REMAINS</a> constant but the pressure and volume of a thermodynamic system will change. The ideal gas equation is ` PV = mu RT` <br/> Work done in an isothermal process: <a href="https://interviewquestions.tuteehub.com/tag/consider-2017521" style="font-weight:bold;" target="_blank" title="Click to know more about CONSIDER">CONSIDER</a> an ideal gas which is allowed to expand quasi-statically at constant temperature from initial state `(P_i , V_i)`to the final state `(P_f , V_f)` . We can calculate the work done by the gas during this process. The work done by the gas, <br/> ` W = int_(v_i)^(v_f) <a href="https://interviewquestions.tuteehub.com/tag/pdv-1149373" style="font-weight:bold;" target="_blank" title="Click to know more about PDV">PDV</a>`....(1) <br/> As the process occurs quasi-statically, at every stage the gas is at equilibrium with the surroundings. Since it is in equilibrium at every stage the ideal gas law is valid . Writing pressure in terms of volume and temperature ,<br/> ` P = (mu RT)/(V )`....(2) <br/>Substituting equation (2) in ( 1) we get<br/>` W =int_(v_i)^(v_f) (mu RT)/(V ) dV` <br/> ` W = mu RT int_(v_i)^(v_f) (dV)/(V) `.....(3) <br/> In equation (3), we take µRT out of the integral, since it is constant throughout the isothermal process. By performing the integration in equation (3 ), we get<br/> `W = mu RT ln (V_f/V_i)`....(4) <br/>Since we have an isothermal expansion , ` (V_f)/(V_i) lt 1` , so ` ln (V_f/V_i) lt 0` . As a result the work done · by the gas during an isothermal expansion is positive . <br/> The above result in equation ( 4) is true for isothermal compression also. But in anisothermal compression `(V_f)/(V_i) lt 1 `, so `ln (V_f/V_i) lt 0` <br/>As a result the work done on the gas in an isothermal compression is negative. In the PV diagram the work done during the isothermal expansion is equal to the area under the graph. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_SP_02_E01_039_S01.png" width="80%"/> <br/>Similarly for an isothermal compression, the area under the PVgraph is equal to the work done on the gas which turns out to be the area with a negative sign.</body></html>


Discussion

No Comment Found

Related InterviewSolutions