\(1 \mathrm{~mm}^{3}\) Of a gas is compressed at 1 atmospheric pressure and temperature \(27^{\circ} \mathrm{C}\) to \(627^{\circ} \mathrm{C}\) What is the final pressure under adiabatic condition. \(\mathrm{r}=1.5\) (A) \(80 \times 10^{5}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (B) \(36 \times 10^{5}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (C) \(56 \times 10^{5}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (D) \(27 \times 10^{5}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\)

First, we need to convert the given temperatures in Celsius to Kelvin using the following formula: \[T(K) = T(^{\circ}\mathrm{C}) + 273.15\] Now, convert the initial and final temperatures: \[T_1 = 27^{\circ}\mathrm{C} + 273.15 = 300.15 \mathrm{K}\] \[T_2 = 627^{\circ}\mathrm{C} + 273.15 = 900.15 \mathrm{K}\]

Using the ideal gas law and making a ratio between initial and final states: \[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\] Since we are not given the volume \(V_2\), we have to express \(V_2\) in terms of \(V_1\) using the adiabatic compression formula: \[P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \Rightarrow V_2^{\gamma - 1} = \frac{P_1}{P_2} V_1^{\gamma}\] Now, substitute this expression for \(V_2\) into our ratio from the ideal gas law: \[\frac{P_1 V_1}{T_1} = \frac{P_2 \left(\frac{P_1}{P_2} V_1^{\gamma}\right)^{\frac{1}{\gamma - 1}}}{T_2}\]

Now we have only one unknown, \(P_2\), and we can solve for it by simplifying the equation in Step 2: \[\frac{P_1}{P_2} = \frac{T_1}{T_2} \left(\frac{P_1}{P_2} V_1^{\gamma}\right)^{\frac{1}{\gamma - 1}}\] Plug in the values of \(T_1\), \(T_2\), and \(\gamma\): \[\frac{P_1}{P_2} = \frac{300.15}{900.15} \left(\frac{P_1}{P_2} (1\mathrm{~mm}^3)^{1.5}\right)^{\frac{1}{0.5}}\] Note that the initial pressure \(P_1\) is given as 1 atmospheric pressure which is equal to \(1 \times 10^5 \mathrm{~N/m^2}\): \[\frac{1 \times 10^5}{P_2} = \frac{300.15}{900.15} \left(\frac{1 \times 10^5}{P_2} (1\mathrm{~mm}^3)^{1.5}\right)^{\frac{1}{0.5}}\] Now solve this equation for \(P_2\) to find the final pressure of the gas: \[P_2 \approx 36 \times 10^5 \mathrm{~N/m^2}\] Hence, the final pressure of the gas under adiabatic condition is approximately \(36 \times 10^5 \mathrm{~N/m^2}\), which is the correct answer (B).