Answer
(a) $n = 0.681 mol$
(b) $V_a = 0.0333 m^3 $
(c) $W = 2.23 kJ$
(d) $\Delta U = 0 $
Work Step by Step
(a) Moles of the gas. We use point b as the reference point.
$n = \frac{p_bV_b}{RT_b} $
$n = \frac{(0.200 atm) (1.013 \times 10^5 Pa/atm)(0.100 m^3)}{(8.314 J/mol.K))(358K)} $
$n = 0.681 mol$
(b) Volume at a when n, R, and T are constant
$V_a = V_b\frac{p_b}{p_a}$
$V_a = (0.10 m^3) \frac{0.200 atm}{0.600 atm}$
$V_a = 0.0333 m^3 $
(c) Work done from a to b
$W = nRTln (\frac{V_b}{V_a})$
$W = (0.681) ((8.314 J/mol.K) (358 K) ln (\frac{0.100 m^3}{0.0333 m^3})$
$W = 2228 J = 2.23 kJ$
(d) The change of internal energy
$\Delta U = n (\frac{3}{2} R)\Delta T$
In isothermal process, $\Delta T = 0$ So
$\Delta U = n (\frac{3}{2} R)(0)$
$\Delta U = 0 $. There is no change of internal energy during this process.
