Answer
(a) $T_a =535 K$
$T_b =9350 K$
$T_c =15000 K$
(b) $W = 21000 J$
(c) $Q = 36000 J$
Work Step by Step
(a) The temperature at point a
$T_a = \frac{p_aV_a}{nR}$
$T_a = \frac{(2.0 \times 10^5 Pa)(0.010 m^3)}{(0.450)(8.314 J/mol.K)}$
$T_a =535 K$
The temperature at point b
$T_b = \frac{(5.0 \times 10^5 Pa)(0.070 m^3)}{(0.450)(8.314 J/mol.K)}$
$T_b =9350 K$
The temperature at point c
$T_c = \frac{(8.0 \times 10^5 Pa)(0.070 m^3)}{(0.450)(8.314 J/mol.K)}$
$T_c =15000 K$
(b) The work done is the area under the graph, the trapezium
$W = \frac{1}{2} (p_a + p_b)(V_b - V_a)$
$W = \frac{1}{2} ((2.0 \times 10^5 Pa + 5.0 \times 10^5 Pa)(0.070 m^3 - 0.010 m^3))$
$W = 21000 J$
(c) The internal energy is 15000 J
$ΔU = Q −W$
$Q = ΔU + W$
$Q = 15000 J +21000 J$
$Q = 36000 J$
