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One mole of ideal monatomic gas is carried through the reversible cyclic process as shown in figure. Calculate the max temperature attained by the gas during the cycle. |
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Answer» `(25)/(8)((P^(@)V^(@))/(R))` `(y-y_(1)) = (Y_(2)-Y_(1))/(x_(2)-x_(1))(x-x_(1))` `(P-3P^(@))=(P^(@)-3P^(@))/(2V^(@)-V_(@))(V-V^(@)) implies (P-3P^(@)) =-(2P^(@))/(V^(@))(V-V^(@)) ""…..(1)` Replacing P by using `P=(RT)/(V)` in equation (1) `((RT)/(V)-3P^(@))=-(2P^(@))/(V^(@)) (V-V^(@)) implies T=(2P^(@)V(V-V^(@)))/(V^(@)R) + (3P^(@)V)/(R)` For T to be maximum `(dT)/(DV)=0` `(dT)/(dV) =-(2P^(@)(2V-V^(@)))/(V^(@)R)+ (3P^(@))/(R) =0 implies V=(5)/(4)V^(@)` `T_("max")=-(2P^(@)xx(5)/(4)V^(@)((5)/(4)V^(@)-V^(@)))/(V^(@)R)+(3P^(@)(5)/(4)V^(@))/(R)=(25)/(8)((P^(@)V^(@))/(R))` |
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