Answer
See explanation.
Work Step by Step
(a) To know the heat transfer, Q, we need to first identify the $\Delta U $ and $W$ .
W is the area under the p versus V curve.
$W = \frac{1}{2} (1.0 ×10^5 Pa + 3.5 ×10^5 Pa)(0.0060 m^3 − 0.0020 m^3 ) + \frac{1}{2}(1.0 ×10^5 Pa + 3.5 ×10^5 Pa)(0.0100 m^3 − 0.0060 m^3 )$
$W = 1800 J $
To find $\Delta U$ we need to identify the changes in temperature of the gasses.
$\Delta T = T_c - T_a = \frac{p\Delta V}{nR}$
$\Delta T = \frac{(1.0 ×10^5 Pa) (0.01 m^3 - 0.002m^3)}{(1/3)(8.314 J/mol.K)}$
$\Delta T = 289 K$
$\Delta U = nC_V \Delta T$
$\Delta U = (1/3)(12.47 J/mol ⋅K)(289 K)$
$\Delta U = 1201 J$
$Q = ΔU +W$
$Q = 1201 J + 1800 J $
$Q = 3001 J$ The heat is transferred into the gas.
(b) Step a to c is isobaric which means the pressure is kept constant. We can straight away use the equation
$Q = nC_p \Delta T$
$Q = (1/3)(20.79 J/mol ⋅K)(289 K)$
$Q = 2002 J$
The heat is transferred into the gas because Q > 0
(c) Q is larger in part (a) than part (b). This means Q in closed loop like stated in part (a) has more heat energy in the system and did more work. It is expected to be more efficient.