Question

An ideal gas at \(27 \mathrm{C}\) is Compressed adiabatically, to \(\\{8 / 27\\}\) of its original Volume. If \(\mathrm{v}=(5 / 3)\), then the rise in temperature is (A) \(225 \mathrm{k}\) (B) \(450 \mathrm{~K}\) (C) \(375 \mathrm{~K}\) (D) \(405 \mathrm{~K}\)

Short answer

The rise in temperature is \(375\,K\), which corresponds to the choice (C).

Step-by-step solution

Write down the given variables

: Initial temperature, \(T_1\) = 27°C = 273 + 27 = 300 K Initial volume, \(V_1\) Final volume, \(V_2\) = 8/27 \(V_1\) Adiabatic Index, \(v\) = 5/3

Write down the equation for the adiabatic process

: We will use the equation for an adiabatic process of an ideal gas: \(\frac{T_1}{T_2} = \left(\frac{V_2}{V_1}\right)^{v-1}\)

Substitute given values into the equation

: We now substitute the given values into the equation: \(\frac{300}{T_2} = \left(\frac{8V_1 / 27V_1}{V_1}\right)^{5/3 - 1}\)

Solve for \(T_2\)

: Simplify the equation and solve for \(T_2\): \(\frac{300}{T_2} = \left(\frac{8}{27}\right)^{2/3}\) \(T_2 = 300 \times \left(\frac{27}{8}\right)^{2/3}\) \(T_2 = 300(1.5)^2\) Therefore, \(T_2 = 675\,K\)

Calculate the rise in temperature

: Now, we just need to find the difference between \(T_2\) and \(T_1\): Temperature rise = \(T_2 - T_1\) Temperature rise = \(675\,K - 300\,K\) Temperature rise = \(375\,K\) The rise in temperature is 375 K, which corresponds to the choice (C).