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Four moles of a diatomic gas (g = 1.4) at a temperature of 300 K were cooled isochorically, so that the final pressure was 1/3 times the original pressure. The gas was allowed to expand isobarically till its temperature got back to its original value. Calculate the total amount of heat absorbed during the process. R = 8.31 J mol ^(-1) K ^(-1).

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/numbe-1126125" style="font-weight:bold;" target="_blank" title="Click to know more about NUMBE">NUMBE</a> of <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> `n = 4` <br/> `gamma = <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.4` <br/> Initial temperature `T_(1) = 300 K` <br/> At the end of the isochoric process the pressure becomes three times the initial pressure. So the temperature becomes `(T_(1))/( 3).` <br/> Heat absorbed during the process is `Q_(1).` <br/> `Q_(1) = Delta <a href="https://interviewquestions.tuteehub.com/tag/u-1435036" style="font-weight:bold;" target="_blank" title="Click to know more about U">U</a> _(1)+ P Deltaa V, Delta V =0` for an isochoric process. <br/> `<a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a> _(1) = (n RT Delta T)/( gamma -1) = ( n R)/( gamma -1) ((T_(1))/( 3) - T_(1))` <br/> `= ( n RT_(1))/( gamma -1)((1-3)/(3))` <br/> `= ( 4 xx 8.31 xx 300 xx ((-2)/(3)))/(1.4 - 1) = - 16620J`</body></html>


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