**(1)**

**In the cyclic process shown in the PV diagram for an ideal gas, the net work done by the gas during one cycle is**

(a) 24P

(a) 24P

_{0}V_{0}(b) –24P_{0}V_{0}(c) – 6πP_{0}V_{0}(d) 6πP_{0}V_{0}(e) 4πP_{0}V_{0}The area enclosed by the closed curve gives the work done. Since the cycle of operations is clockwise (in the PV diagram with P on the Y-axis), the net work done by the gas is positive

Therefore net work done by the system during one cycle = Area of circle.

Since the quantities on the two axes are different, the shape is not rally a circle. The scales of the quantities could have been changed to make it an ellipse. So, treat it as an ellipse and calculate the area as πab where ‘a’ and ‘b’ are the semi-major and semi-minor axes respectively.

So, work done = π[(5P

[Note that if the arrow in the diagram shows the processes in the anticlockwise direction, the net work done by the system will be negative].

Therefore w = Q/β = 1000/10 = 100 calories = 420 J, approximately. This is the minimum work to be done by the compressor motor.

Actual efficiency = (10000–8000)/10000 = 1/5.

Since the quantities on the two axes are different, the shape is not rally a circle. The scales of the quantities could have been changed to make it an ellipse. So, treat it as an ellipse and calculate the area as πab where ‘a’ and ‘b’ are the semi-major and semi-minor axes respectively.

So, work done = π[(5P

_{0}–2P_{0})/2] [(10V_{0}–2V_{0})/2] = 6πP_{0}V_{0}[Note that if the arrow in the diagram shows the processes in the anticlockwise direction, the net work done by the system will be negative].

**(2) A refrigerator removes 1000 calories of heat from the ice trays. The coefficient of performance of the refrigerator is 10. Then the work done by the compressor motor is at least**

(a) 240 J (b) 420 J (c) 4200 J (d) 42000 J (e) 0.042 J

Coefficient of performance, β = Q2/w where Q(a) 240 J (b) 420 J (c) 4200 J (d) 42000 J (e) 0.042 J

_{2}is the quantity of heat removed from the cold body (inside the refrigerator) and ‘w’ is the work needed to transfer this heat to the hot body (outside).Therefore w = Q/β = 1000/10 = 100 calories = 420 J, approximately. This is the minimum work to be done by the compressor motor.

**(3) An engine takes in 10000 J of heat and rejects 8000 J while operating between temperatures of 900 K and 600 K. The actual efficiency is**

(a) 1/5 (b) 2/5 (c) 1/3 (d) 2/3 (e)4/5

This simple question appeared in the Kerala Engineering Entrance 2000 test paper. The actual efficiency is given by η = (Q(a) 1/5 (b) 2/5 (c) 1/3 (d) 2/3 (e)4/5

_{1}–Q_{2})/Q_{1}. Note that this will be less than (T_{1}–T_{2})/T_{1}in practical heat engines. The efficiency given by the two expressions will be equal in the case of the ideal, perfectly reversible Carnot engine only.Actual efficiency = (10000–8000)/10000 = 1/5.