Answer
See explanation.
Work Step by Step
(a) Initial temperature is
$(T_1V_1)^{\gamma - 1} = (T_2V_2) ^{\gamma - 1} $ and $pV = nRT$
$T_1 = \frac{pV}{nR} $
$T_1 = \frac{(1.00\times 10^5 Pa)(2.50\times 10^{-3} m^3) }{(0.1 mol)(8.3145 J/mol.K)}$
$T_1 = 301K$
(b) Final temperature if the gas is
i. Isothermal means constant temperature, so the final and initial temperatures are the same which is $T_2 = 301K$
The pressure of the gas is
$p_2= p_1(\frac{V_1}{V_2}) $
$p_2= \frac{1}{2} p_1$
$p_2= \frac{1}{2} (1.0 \times 10^5 Pa)$
$p_2= 5.0 \times 10^4 Pa$
ii. Isobaric means constant pressure, so the pressure is
$\Delta p = 0$ so $p_1 = p_2$
$p_2 = 1.0 \times 10^5 Pa)$
For the temperature, we use
$T_2= T_1(\frac{V_2}{V_1}) $
$T_2= 2T_1 $
$T_2 = 2(301K)$
$T_2 = 602 K$
iii. Adiabatic. We use
$T_2= \frac{T_1V_1^{\gamma - 1}}{V_2^{\gamma - 1}} $
$T_2= \frac{(301K)V_1^{0.67}}{2.0V_1^{0.67} }$
$T_2= (301K)(\frac{1}{2.0})^{0.67} $
$T_2= 189 K$
The pressure can be calculated by
$pV = nRT$
$p= \frac{(0.1mol)(8.3145 J/mol.K)(189K)}{(2 \times 2.50 \times 10^{-3} m^3) } $
$p = 3.14 \times 10^4 Pa$
