Question

A heat engine with 0.500 mol of a monatomic ideal gas initially fills a 3000 cm3 cylinder at 800 K. The gas goes through the following closed cycle: - Isothermal expansion to 4000 cm3. - Isochoric cooling to 300 K. - Isothermal compression to 3000 cm3. - Isochoric heating to 800 K.

Answer 1

Answer:

Explanation:

$\text{For isothermal expansion:}$

$W_1 = nRT_1 In (\dfrac{V_2}{V_1}) \\ \\ W_1 = 0.5 \times 8.314 \times 800 \times In (\dfrac{4000}{3000})\\ \\ W _1 = 956.72 \ J \\ \\ Q_1 = W_1 = 956.72 \ J \ since \ (dU=0) \\ \\ \\ \text{For isochoric cooling,} W_2 = 0} \\ \\ Q_2 = nCr \Delta T \\ \\ Q_2 = 0.5 (\dfrac{3R}{2})(T_2-T_1) \\ \\ Q_2 = 0.5 \times \dfrac{3\times 8.314}{2}(-500)= -3117.75$

$\text{For Isothermal compression:}\\\\ W_3 = nRT_2 \ In (\dfrac{V_4}{V_3}) \\ \\ 0.5 \times 8.314 \times 300 \times In (\dfrac{3000}{4000}) \\ \\ W_3 =- 358.77 \ J \\ \\ Q_3=W_3= - 358.77 \ J$

$\text{For isochoric heating; }W_4 =0} \\ \\ Q_4 = nC_v\Delta T \\ \\ = 0.5 \times \dfrac{3}{2}\times 8.314 \times 500 \\ \\ Q_4 = 3117.5 \ J$

$\text{Total workdone W}= W_1 + W_2+W_3+W_4 \\ \\ W = 956.71 \ J + 0 + (-358.77 \ J) +0 \\ \\ \mathbf{W = 597.94 J} \\ \\ \\ \eta = \dfrac{Work \ done}{heat \ taken } \\ \\ \eta = \dfrac{W}{Q_1+Q_4} \\ \\ \eta = \dfrac{597.94 \ J}{956.71 \ J + 3117.5 \ J} \\ \\ \eta = 0.1468 \\ \\ \mathbf{\eta = 14.68\%}$