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Find the maximum attainable temperature of ideal gas in each of the following process : (a) `p = p_0 - alpha V^2` , (b) `p = p_0 e^(- beta v)`, where `p_0, alpha` and `beta` are positive constants, and `V` is the volume of one mole of gas. |
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Answer» Correct Answer - (a) `T_(max) = (2)/(3)(p_(0)//R) sqrt(p_(0)//3 alpha)]` (b) `T_(max) = p_(0)//e betaR`. (a) `pV = RT` `V = (RT)/(p)` `p = p_(0) - alphaV^(2)` `p = p_(0) - alpha ((RT)/(p))^(2)` `T = (1)/(Rsqrt(alpha)) sqrt(p_(0)p^(2)-p^(3))` for maximum value `T, sqrt(p_(0)p^(2)-p^(3))` should be maximum `(d)/(dP) (p_(0)p^(2)-p^(3)) = 0` `p = (2p_(0))/(3)` `T = (1)/(Rsqrt(alpha)) sqrt(p_(0)p^(2)-p^(3))` `p =(2p_(0))/(3)` `T_(max)= (2)/(3) (p_(0))/(R ) sqrt((p_(0))/(3alpha))` (b) `p = p_(0)e^(-betaV)` or `p = p_(0) e^(-(betaRT)/(p))` for maximum value of `T` `(dT)/(dp) = 0` `p = p_(0)e^(-(betaRT)/(p))` `ln (p) = ln p_(0) -(betaRT)/(p)` `ln ((p)/(p_(0))) =- beta (RT)/(p)` `T =- (p)/(betaR) ln ((p)/(p_(0)))` for `T_(max)` After solving `p = (p_(0))/(e) rArr T_(max) = (p_(0))/(e betaR)`. |
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