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An ideal gas at \mathrm{127^{\circ} \mathrm{C}} is compressed adiabatically to onetenth of its volume. If the gas obeys the equation of state \mathrm{ P V{\frac{4}{3}}=} constant, the percentage decrease in the gas temperature will be:

Option: 1

40


Option: 2

50


Option: 3

60


Option: 4

70


Answers (1)

For an adiabatic process, the relationship between pressure (P), volume (V), and adiabatic index \mathrm{(\gamma)} is given by: \mathrm{P V^\gamma=}constant.

In this case, the initial temperature is \mathrm{127^{\circ} \mathrm{C}}, which is \mathrm{127+273.15=400.15}
K. The initial volume is V, and the final volume is \mathrm{\frac{V}{10}.}
Since the gas obeys the equation of state \mathrm{P V{\frac{4}{3}}=} constant, we have:

                                                \mathrm{ P_i \cdot V_i{\frac{4}{3}}=P_f \cdot V_f{\frac{4}{3}} }
Where \mathrm{ P_i } and \mathrm{ P_f} are the initial and final pressures, and \mathrm{ V_i} and \mathrm{ V_f} are the initial and final volumes, respectively.

Since the process is adiabatic, \mathrm{ P V^\gamma=} constant, where \mathrm{ \gamma=\frac{4}{3}}. So, \mathrm{ P_i \cdot V_i^{\frac{4}{3}}=P_f \cdot V_f^{\frac{4}{3}}} becomes:

                                              \mathrm{ \begin{gathered} P_i \cdot V_i^{\frac{4}{3}}=P_f \cdot\left(\frac{V_i}{10}\right)^{\frac{4}{3}} \\ \frac{P_i}{P_f}=10^{\frac{4}{3}} \end{gathered} }

Now, let's find the final temperature \mathrm{\left(T_f\right)} after compression:

                                        \mathrm{ \begin{gathered} \frac{T_i}{T_f}=\left(\frac{P_i}{P_f}\right)^{\frac{\gamma-1}{\gamma}} \\\\ \frac{T_i}{T_f}=10^{\frac{4 / 3-1}{4 / 3}}=10^{\frac{1}{4}} \\\\ T_f=\frac{T_i}{10^{\frac{1}{4}}}=\frac{400.15}{\sqrt[4]{10}} \end{gathered} }
The percentage decrease in temperature is given by:

\mathrm{ \begin{gathered} \text { Percentage decrease }=\frac{T_i-T_f}{T_i} \times 100 \\ \text { Percentage decrease }=\frac{400.15-\frac{400.15}{\sqrt[4]{10}}}{400.15} \times 100 \end{gathered} }

So, the correct answer is approximately: 1) \mathrm{\mathbf{4 0} \%}

Posted by

Ramraj Saini

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