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Question Number 58196 by Umar last updated on 19/Apr/19

$$\mathrm{The}\mathrm{molar}\mathrm{heat}\mathrm{capacity}\mathrm{of}\mathrm{constant}$$$$\mathrm{presure}\mathrm{of}\mathrm{a}\mathrm{gas}\mathrm{varies}\mathrm{with}\mathrm{the}\mathrm{temperature}$$$$\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{equation}$$$${\mathrm{C}}_{\mathrm{p}}=\mathrm{a}+\mathrm{b}\theta -\frac{\mathrm{C}}{{\theta }^2}$$$$\mathrm{where}\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{C}\mathrm{are}\mathrm{constants}.$$$$\mathrm{How}\mathrm{much}\mathrm{heat}\mathrm{is}\mathrm{transfered}\mathrm{during}$$$$\mathrm{an}\mathrm{isobaric}\mathrm{process}\mathrm{in}\mathrm{which}\mathrm{n}\mathrm{mole}$$$$\mathrm{of}\mathrm{gas}\mathrm{undergo}\mathrm{a}\mathrm{temperature}\mathrm{rise}$$$$\mathrm{from}{\theta }_i\mathrm{to}{\theta }_f?$$

Answered by tanmay last updated on 19/Apr/19

$$dQ={nC}_{p}d\theta$$$$dQ=n(a+b\theta -\frac{C}{{\theta }^2})d\theta$$$$Q=na{\int }_{{\theta }_i}^{{\theta }_f}d\theta +nb{\int }_{{\theta }_i}^{{\theta }_f}\theta d\theta -nC{\int }_{{\theta }_i}^{{\theta }_f}{\theta }^{-2}d\theta$$$$=na\mid \theta {\mid }_{{\theta }_i}^{{\theta }_f}+nb\mid \frac{{\theta }^2}{2}-{\mid }_{{\theta }_i}^{{\theta }_f}-nC\mid \frac{{\theta }^{-1}}{-1}{\mid }^{{\theta }_f}{\theta }_i$$$$=na({\theta }_f-{\theta }_i)+\frac{nb}{2}({\theta }_f^2-{\theta }_i^2)+nC(\frac{1}{{\theta }_f}-\frac{1}{{\theta }_i})$$$$plzcheck...$$$$$$

Commented by Umar last updated on 19/Apr/19

$$\mathrm{thanks}\mathrm{a}\mathrm{bunch}$$

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