A gas expands with temperature according to the relation V = KT^(2//3) where K is a constant. The work done by the gas when the temperature changes by 60 K is

A gas expands with temperature according to the relation V = KT^(2//3)

1.	A gas expands with temperature according to the relation V = KT^(2//3) where K is a constant. The work done by the gas when the temperature changes by 60 K is
Answer» <html><body><p>10 R<br/>30 R<br/>40 R<br/>20 R</p>Solution :`dW = PdV = (RT)/(V)<a href="https://interviewquestions.tuteehub.com/tag/dv-433533" style="font-weight:bold;" target="_blank" title="Click to know more about DV">DV</a>` …(i)<br/> As `V = KT^(2//<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) :. dV = K(2)/(3)T^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>//3) dT` <br/>`:. (dV)/(V) = (K(2)/(3)T^(-1//3)dT)/(KT^(2//3)) = (2)/(3) (dT)/(T)` ….(<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) <br/> From (i), `W = underset(T_(1))overset(T_(2))int RT(dV)/(V) = underset(T_(1))overset(T_(2))int RT (2)/(3)(dT)/(T) = (2)/(3)R[T]_(T_(1))^(T_(2))` (Using (ii)) <br/> `= (2)/(3)R(T_(2) - T_(1)) = (2)/(3) R xx <a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a> = 40R`</body></html>