Question

# An ideal gas is taken through a cyclic thermodynamic process through four steps. The mass of heat involved in these steps are: Q1 = 5960J, Q2 = -5585J, Q3 = -2980J and Q4 = 3645J, respectively. The corresponding quantities of work involved are w1 = 2200J, w2 = -825J and w3 = 1100J respectively. The value of w4 (in J) is:(Write the absolute value of an answer)

Thermodynamic process in which the system changes from one thermodynamic state to the final stat due to change in the temperature, pressure, volume, etc. According to the first law of thermodynamics, $\text{Q = }\Delta \text{U + W}$, where q is the heat absorbed by the system, $\Delta \text{U}$is the change in the internal energy and W is the work done by the system.

-The first law of thermodynamics states that the energy can neither be created nor destroyed by anyone and hence, it is conserved but it can be converted from one form to another form.
-For cyclic process, the value of $\Delta \text{U}$ is zero because in the cyclic process the initial concentration and final concentration is the same.
-So, the formula given according to the first law of thermodynamics will be:
\begin{align} & \text{Q = }\Delta \text{U + W (as }\Delta \text{U}=0) \\ & \text{Q = W }.....\text{(1)} \\ \end{align}
-It is given that the value of ${{\text{Q}}_{1}},\text{ }{{\text{Q}}_{2}}\text{, }{{\text{Q}}_{3}}\text{ and }{{\text{Q}}_{4}}$is 5960, 5585, 2980 and 3645 so the net heat absorbed will be:
Q = 5960 - 5585 - 2980 + 3645 = 1040J
-Similarly, the total work will be: $\text{W = }{{\text{W}}_{1}}\text{ + }{{\text{W}}_{2}}\text{ + }{{\text{W}}_{3}}\text{ + }{{\text{W}}_{4}}$
W = 2200 - 825 - 1100 + ${{\text{W}}_{4}}$ …. (2)
-According to the equation (1), W is equal to Q so the equation (2) will become:
1040 = 275 + ${{\text{W}}_{4}}$
So, ${{\text{W}}_{4}}$ = 1040 -275 = 765 Joules.
Therefore, the work done by the system in the fourth step will be 765 Joules.

Note: The efficiency of a system can also be calculated by using the ratio of work done and heat absorbed i.e. $\text{Efficiency = }\dfrac{\text{Q}}{{{\text{W}}_{1}}\text{ + }{{\text{W}}_{4}}}$$\text{= }\dfrac{1040}{9605}\text{ = 0}\text{.1082}$. So, the efficiency will be 10.82% (by multiplying 100). It means in this process only 10.82% useful energy is converted from the system.