An ideal gas is taken through cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are Q1 = 5970 J, Q2 = -5485 J, Q3 = -2900 J, Q4 = 3650 J, respectively. The corresponding works involved are W1 = 2500 J, W2 = -725 J, W3 = -1200 J and W4, respectively. The value of W4 is :
Answer & explanation
Correct answer: option 1
\(\Delta Q = Q_1 + Q_2 + Q_3 + Q_4\)
\(\Delta Q = 5970 - 5485 - 2900 + 3650 \) = 1235 J
\(\Delta W = W_1 + W_2 + W_3 + W_4\)
\(\Delta W = 2500 - 725 - 1200 + W_4\) = 575 + W4
For a cyclic process : Uf = Ui
\(\Delta U = U_f - U_i\) = 0
From First Law of Thermodynamics : \(\Delta Q = \Delta U + \Delta W\)
1235 + 0 = 575 + W4
\(\Rightarrow W_4 = 660\) J