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1 liter dry air at STP expands adiabatically to a volume of 3L. If gamma=1.4, the work done by air is (3^(1.4)= 4.655) (Take air to be an ideal gas)

Answer» <html><body><p>18J<br/>45J<br/>90.5J<br/>100.8J</p>Solution :STP means `P_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)= 1` <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a>, T= <a href="https://interviewquestions.tuteehub.com/tag/273k-299107" style="font-weight:bold;" target="_blank" title="Click to know more about 273K">273K</a> and `V_(1)=1L` <br/> After adiabatic expansion, `P_(2)=? And V_(2)= 3L` <br/> For adiabatic expansion <br/> `P_(1)V_(1)^(gamma) = P_(2) V_(2)^(gamma)` [ `:.T` is constant] <br/> `:. P_(2) = P_(1) ((V_(1))/(V_(2)))^(gamma)` <br/> `=1 ((1)/(3))^(gamma)` <br/> `=((1)/(3))^(1.4) [ :. gamma= 1.4]` <br/> `= (1)/(3^(1.4))= (1)/(4.655)` <br/> = 0.2148 atm <br/> <a href="https://interviewquestions.tuteehub.com/tag/work-20377" style="font-weight:bold;" target="_blank" title="Click to know more about WORK">WORK</a> done in adiabatic process, `W= (P_(1)V_(1)- P_(2)V_(2))/(gamma-1)` <br/> `=(1 xx 1-0.2148 xx 3)/(1.4-1)` <br/> `=(1-0.6444)/(0.4)` <br/> `=(0.3556)/(0.4)` <br/> `W= 0.889 xx 1.01 xx 10^(5) xx 10^(-3)<a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a>` 91 atm = `1.01 xx 10^(5) (N)/(m^(2)) 1L= 10^(-3) m^(3)`] <br/> `:. W= 0.89789xx 10^(2)J` <br/> `:. W= 89.789J` <br/> `:. W= 90J`</body></html>


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