An ideal gas sample ( γ = 1.5 ) is adiabatically compressed from a...

For an adiabatic process, the relation between temperature and volume is given by $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.

Substituting the given values:

$T_2 = T_1 \left(\dfrac{V_1}{V_2}\right)^{\gamma - 1} = 300 \left(\dfrac{150}{50}\right)^{1.5 - 1} = 300(3)^{0.5} = 300\sqrt{3} \text{ K}$

The number of moles of the gas is given by $n = \dfrac{P_1 V_1}{R T_1}$.

The change in internal energy is $\Delta U = n C_v \Delta T = \dfrac{n R (T_2 - T_1)}{\gamma - 1}$.

Substituting $n R = \dfrac{P_1 V_1}{T_1}$:

$\Delta U = \dfrac{P_1 V_1}{T_1} \dfrac{(T_2 - T_1)}{\gamma - 1} = \dfrac{P_1 V_1}{\gamma - 1} \left(\dfrac{T_2}{T_1} -1\right)$

Given $P_1 = 150 \times10^3 \text{ Pa}$and $V_1 = 150 \times10^{-6} \text{ m}^3:<br><br>$P_1 V_1 = 150 \times10^3 \times150 \times10^{-6} = 22.5 \text{ J}$<br><br>$\Delta U = \dfrac{22.5}{1.5 - 1} (\sqrt{3} -1) = \dfrac{22.5}{0.5} (1.732 - 1)$<br><br>$\Delta U = 45 \times0.732 = 32.94 \text{ J} \approx33 \text{ J}