Question

A gas is initially at a pressure of 100 kPa and its volume is 2.0 m3. Its pressure is kept constant and the volume is changed from 2.0 m3 to 2.5 m3. Its Volume is now kept constant and the pressure is increased from 100 kPa to 200 kPa. The gas is brought back to its initial state, the pressure varying linearly with its volume. (a) Whether the heat is supplied to or extracted from the gas in the complete cycle? (b) How much heat was supplied or extracted?

Answer 1

(a) The heat is removed from the device as there is zero increase in internal energy.

(b) Heat supplied or extracted is 25000 J.

Explanation:

Step 1:

(a) P$_1$ = 100 kPa,

V$_1$ = 2 $m^3$  

V$_2$ = 2.5 $m^3$

∆V = 0.5 $m^3$

W = P∆V  

W is work done  

$\mathrm{W}=100 \times 10^{3} \times 0.5 \mathrm{W}$

W=100×1000×0.5 W  

W=10×1000×5 W  

W=50×1000 W  

$=5 \times 10^{4} \mathrm{J}$

$\mathrm{W}_{\mathrm{AB}}=\text { Area under } \mathrm{AB} \text { line }=5 \times 10^{4} \mathrm{J}$

Step 2:

If volume is constant for BC  line, then ∆V = 0.

$W_B_C$ = P∆V = 0

Work done at point B to point C = 0

When the system returns from C to the initial point A, the work done is equal to the area under AC line .

$W_C_A$ = triangle area ABC + rectangle Area  under line AB

Net work done, W = enclosed Area by the ABCA

We see from the graph that the region under AC is greater than that under AB.

$\mathrm{W}=\mathrm{W}_{\mathrm{AC}}-\mathrm{W}_{\mathrm{AB}}$

We see also that heat is removed from the device as there is zero increase in internal energy.

Step 3:

(b) Extracted heat = enclosed area under ABCA

$\begin{aligned}&=\frac{1}{2} \times 0.5 \times 100 \times 10^{3}\\&0.5 \times 50 \times 10^{3}\\&5 \times 5 \times 10^{3}\\&25 \times 10^{3}=25000 J\end{aligned}$

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

(a) heat is removed from the device as there is zero internal energy

(b) 25000 joule