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At 500 \mathrm{~K}$ and $150\ \mathrm{kPa}, a given volume of ideal gas 500 \mathrm{~K}$ and $150\ \mathrm{kPa} was compressed adiabatically. What should be the minimum pressure so temperature is raised to 750 \mathrm{~K} ? The compression is adiabatically irreversible.

Option: 1

\mathrm{362.5\ kPa}


Option: 2

\mathrm{287.5\ kPa}


Option: 3

\mathrm{412.5\ kPa}


Option: 4

\mathrm{555.5 \ kPa}


Answers (1)

For an adiabatic irreversible compression,
\mathrm{\begin{gathered} \Delta V=\omega \\ \therefore n C_{V_{1} m}\left(T_2-T_1\right)=-P_{\text {ext }}\left(V_2-V_1\right) \end{gathered} }
\mathrm{here\ P_{\text {ext }}=P_2 ; \quad P_1=100\ \mathrm{kPa}} \\
\mathrm{n C_{v_{1} m}\left(T_2-T_1\right)=-P_2\left[\frac{n R T_2}{P_2}-\frac{n R T_1}{P_1}\right]}
\frac{5}{2} R\left(T_2-T_1\right)=-P_2 R\left[\frac{T_2}{P_2}-\frac{T_1}{P_1}\right]
\frac{5}{2}(750-500)=-B_2\left[\frac{750}{P_2}-\frac{500}{150}\right]
\begin{aligned} \frac{6}{2} \times 250 & =-750+\frac{500}{150} P_2 \\ P_2 & =412.5\ \mathrm{kPa} \end{aligned}

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Kshitij

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