An ideal gas is taken through cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are Q1 = 5960 J, Q2 = -5585 J, Q3 = -2980 J, Q4 = 3645 J, respectively. The corresponding works involved are W1 = 2200 J, W2 = -825 J, W3 = -1100 J and W4, respectively. The value of W4 is :
Answer & explanation
Correct answer: option 3
\(\Delta Q = Q_1 + Q_2 + Q_3 + Q_4\)
\(\Delta Q = 5960 - 5585 - 2980 + 3645 \) = 1040 J
\(\Delta W = W_1 + W_2 + W_3 + W_4\)
\(\Delta W = 2200 - 825 - 1100 + W_4\) = 275 + W4
For a cyclic process : Uf = Ui
\(\Delta U = U_f - U_i\) = 0
From First Law of Thermodynamics : \(\Delta Q = \Delta U + \Delta W\)
1040 + 0 = 275 + W4
\(\Rightarrow W_4 = 765\) J