A gas expand with temperature according to the relation V = KT23. What is the work done when the temperature changes by 300C?

W = ∫PdV = ∫RTVdVSince V = KT23Hence, dV = 23KT-13dt ∴ dVV = 23.dTTHence W = ∫T1T223RTTdT = 23R(T2-T1) = 23R(30) = 20R

A gas expand with temperature according to the relation V = KT23. What is the work done when the temperature changes by 300C?

W = ∫PdV = ∫RTVdVSince V = KT23Hence, dV = 23KT-13dt ∴ dVV = 23.dTTHence W = ∫T1T223RTTdT = 23R(T2-T1) = 23R(30) = 20R

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A gas expand with temperature according to the relation $V = {KT}^{\frac{2}{3}}$ . What is the work done when the temperature changes by $30^{0} C$ ?

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Detailed Solution

$W = \int PdV = \int \frac{RT}{V} dV$

Since V = ${KT}^{\frac{2}{3}}$

Hence, $dV = \frac{2}{3} {KT}^{- \frac{1}{3}} dt ∴ \frac{dV}{V} = \frac{2}{3} . \frac{dT}{T}$

Hence W = $\int_{T_{1}}^{T_{2}} \frac{2}{3} \frac{RT}{T} dT$

= $\frac{2}{3} R (T_{2} - T_{1}) = \frac{2}{3} R (30) = 20 R$

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A gas expand with temperature according to the relation V = KT23. What is the work done when the temperature changes by 300C?

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A gas expand with temperature according to the relation $V = {KT}^{\frac{2}{3}}$ . What is the work done when the temperature changes by $30^{0} C$ ?