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A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion ( in the intermediate stages of expansion/ compression the states of gases are not defined). The work done can be calculated using dw=-P_("ext")dV while in case of reversible process the work done can be calculated using dw=-PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process, since P=(nRT)/(V),so, w=intdw=-int_(V_(i))^(V_(f))(nRT)/(V).dV=-nRT ""In((V_(f))/(V_(i))) Since,dw=PdV, so magnitude of work done can also be calculated by calculating the area under the PV curve of the reversible process in PV diagram. There are two sample of same gas initially at same initial state. Gases of both the sample are expanded. Ist sample using reversible isothermal process and IInd sample using reversible adiabatic process till final pressures of both the samples become half of the initial pressure , then :

Answer» <html><body><p>Final <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of <a href="https://interviewquestions.tuteehub.com/tag/ist-501349" style="font-weight:bold;" target="_blank" title="Click to know more about IST">IST</a> <a href="https://interviewquestions.tuteehub.com/tag/sample-1194587" style="font-weight:bold;" target="_blank" title="Click to know more about SAMPLE">SAMPLE</a> `<a href="https://interviewquestions.tuteehub.com/tag/lt-537906" style="font-weight:bold;" target="_blank" title="Click to know more about LT">LT</a>` final volume of Iind sample<br/>final volume of IInd sample `lt`final volume of Ist sample<br/>final volumes will be equal<br/>information insufficient</p>Answer :b</body></html>


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