If the pressure and the volume of certain quantity of ideal gas are halved, then its temperature\

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Professor PC Thomas Classes · Answer

According to ideal gas law\ $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}} \quad$ or $\quad \mathrm{T}_{2}=\mathrm{T}_{1} \frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{P}_{1} \mathrm{~V}_{1}}$\ Here, $\mathrm{P}_{1}=\mathrm{P}, \mathrm{V}_{1}=\mathrm{V}, \mathrm{T}_{1}=\mathrm{T}$\ $\mathrm{p}_{2}=\frac{\mathrm{p}}{2}, \mathrm{~V}_{2}=\frac{\mathrm{V}}{2}, \mathrm{~T}_{2}=$ ?\ $	herefore \quad \mathrm{T}_{2}=\frac{\mathrm{T}\left(\frac{\mathrm{P}}{2}ight)\left(\frac{\mathrm{V}}{2}ight)}{\mathrm{pV}}$ or $\mathrm{T}_{2}=\frac{\mathrm{T}}{4}$\