Question

Figure shows three paths through which a gas can be taken from the state A to the state B. Calculate the work done by the gas in each of the three paths.
Figure

Answer 1

Three paths through which a gas can be taken from the state A to the state B. The work done by the gas in each of the three paths = 0.15 J

Explanation:

Step 1:

W  = PΔV  

W is all the work done,

If during a process pressure and volume switch, then you can find out what work is done by finding the region Below diagram of PV.

In ACB path, at line AC:

As the initial volume is the same as the final volume, ΔV=0

$W_{\mathrm{AC}}=\mathrm{P} \Delta \mathrm{V}=0$

Step 2:

at BC line:

$P = 30 \times10^3 \,Pa$

$W_{ACB} = W_{AC} + W_{BC} = 0 + P\Delta V$

$= 30 \times 10^3 \times (25-10) \times 10^{-6}$

$= 30 \times 10^3 \times 15 \times 10^{-6}$

$= 450 \times 10^3 \times 10^{-6}$

= 0.45 J

Step 3:

For AB path:

If both volume and pressure change, the medium pressure is used to get the work done.

Pressure (P) =  $\frac{1}{2}\times(30+10)\times 10^3$

$W_{AB} = \frac{1}{2} \times (30+10)\times10^3\times 15 \times 10^{-6}$

$= \frac{1}{2} \times 40 \times 15 \times10^{-3} = 0.30\, J$

Step 4:

Initial volume in ADB path, along DB line is identical to the final volume. hence, the work done along that line is zero.

In route  AD, P = 10 kPa

$W = W_{AD} + W_{DB}$

$= 10 \times 10^3 (25-10) \times 10^{-6} + 0$

$= 10 \times 15 \times 10^{-3} = 150 \times 10^{-3}$

= 0.15 J

Answer 2

Answer:

Three paths through which a gas can be taken from the state A to the state B. The work done by the gas in each of the three paths = 0.15 J

Explanation:

process 1:  W  = PΔV    

W is all the work done,

If during a process pressure and volume switch, then you can find out what work is done by finding the region Below diagram of PV.  

In ACB path, at line AC:  As the initial volume is the same as the final volume, ΔV=0

process 2:

at BC line:

= 0.45 J

For AB path:

If both volume and pressure change, the medium pressure is used to get the work done.

Pressure (P) =  

process 4:

Initial volume in ADB path, along DB line is identical to the final volume. hence, the work done along that line is zero.

In route  AD, P = 10 kPa

= 0.15 J

Explanation: