One mole of an ideal gas left( {frac{{{Cₚ}}}{{{Cᵥ}}} = Y} right) heated by law P = ?

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11.Thermodynamics

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One mole of an ideal gas $\left( {\frac{{{C_p}}}{{{C_v}}} = Y} \right)$ heated by law $P = \alpha V$ where $P$ is pressure of gas, $V$ is volume, $\alpha $ is a constant. What is the molar heat capacity of gas in the process

$C = \frac{R}{{\gamma - 1}}$

$C = \frac{{\gamma R}}{{\gamma - 1}}$

$C = \frac{R}{2}\frac{{\left( {\gamma - 1} \right)}}{{\left( {\gamma + 1} \right)}}$

$C = \frac{R}{2}\frac{{\left( {\gamma + 1} \right)}}{{\left( {\gamma - 1} \right)}}$

Solution

$\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{n} \Delta \mathrm{T}} \frac{\Delta \mathrm{U}+\mathrm{W}}{\mathrm{n} \Delta \mathrm{T}}=\frac{\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{t}+\int_{\mathrm{V}_{f}}^{\mathrm{V}_{\mathrm{i}}} \mathrm{PdV}}{\mathrm{n} \Delta \mathrm{T}}$

$=\mathrm{C}_{\mathrm{V}}+\frac{\alpha}{\mathrm{n} \Delta \mathrm{t}} \cdot \int_{\mathrm{V}_{\mathrm{i}}}^{\mathrm{V}_{\mathrm{f}}} \mathrm{VdV}$

$=\frac{R}{\gamma-1}+\frac{1}{2} \frac{\alpha V_{f}^{2}-\alpha V_{i}^{2}}{n \Delta T}$

$=\frac{\mathrm{R}}{\gamma-1}+\frac{1}{2} \frac{\mathrm{P}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}}{\mathrm{n} \Delta \mathrm{T}}$

$=R\left[\frac{1}{\gamma-1}+\frac{1}{2}\right]=\frac{R}{2}\left(\frac{\gamma+1}{\gamma-1}\right)$

Standard 11

Physics

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One mole of an ideal gas $\left( {\frac{{{C_p}}}{{{C_v}}} = Y} \right)$ heated by law $P = \alpha V$ where $P$ is pressure of gas, $V$ is volume, $\alpha $ is a constant. What is the molar heat capacity of gas in the process

Solution

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