Answer
(a)$T_a = 278.5 K$
(b)$W_{ab} = 0 $
$W_{bc} = 162 J $
(c)$ΔU = 53 J $
Work Step by Step
(a) The lowest temperature is at the point (a). We know it because it has the lowest pressure and volume.
$T_a = \frac{p_aV_a}{nR}$
$T_a = \frac{(0.20 atm)(1.013 \times 10^5 Pa/atm)(2.0 L)(1.0 \times 10^{-3} m^3/L)}{(0.0175)(8.314 J/mol.K)}$
$T_a = 278.5 K$
(b) Work done from a to b
$W_{ab} = p(\Delta V)$ and in the figure, there is no change in volume so $\Delta V = 0$
$W_{ab} = p(0)$
$W_{ab} = 0 $
Work done from b to c, area under the curve is
$W_{bc} = \frac{1}{2} (0.50 atm + 0.30 atm)(6.0 L − 2.0 L) $
$W_{bc} = (1.6 atm. L) (1.013 \times 10^5 Pa/atm)(1.0 \times 10^{-3} m^3/L)$
$W_{bc} = 162 J $
(c) The internal energy is
$ΔU = Q −W$
$ΔU = 215 J - 162 J $
$ΔU = 53 J $
