For the thermodynamic process shown in the figure- PA - 1 - 105 Pa- PB - 0-3 - 105 Pa PD - 0-6 - 105 Pa- VA - 0-20 litre VD - 1-30 litreFind the work performed in joules by the system along path AD-

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Standard XII

Chemistry

Internal Energy

For the therm...

Question

For the thermodynamic process shown in the figure.
$P_{A} = 1 \times 10^{5} P a; P_{B} = 0.3 \times 10^{5} P a$
$P_{D} = 0.6 \times 10^{5} P a; V_{A} = 0.20 l i t r e$
$V_{D} = 1.30 l i t r e$
Find the work performed (in joules) by the system along path $A D$ .

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Solution

Work done along $A D$ is equal to the area under the line $A D$ .
Let E be the intersection between $A B$ and straight line through $D$ .
Let G and F be the points on $x$ axis completing the rectangle EDFG
Area under $A D =$ Area of triangle $A D E +$ Area of rectangle $E D F G$
$= \frac{1}{2} \times A E \times E D + E G \times E D$
$= \frac{1}{2} \times (P_{A} - P_{D}) \times (V_{D} - V_{A}) + (P_{D} - 0) \times (V_{D} - V_{A})$
$= \frac{1}{2} \times 0.4 \times 10^{5} \times 1.1 \times 10^{- 3} + 0.6 \times 10^{5} \times 1.1 \times 10^{- 3}$
$= 22 + 66 = 88 J$

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For the thermodynamic process shown in the figure. PA=1×105Pa;PB=0.3×105PaPD=0.6×105Pa;VA=0.20litre VD=1.30litreFind the work performed (in joules) by the system along path AD.

For the thermodynamic process shown in the figure.
$P_{A} = 1 \times 10^{5} P a; P_{B} = 0.3 \times 10^{5} P a$
$P_{D} = 0.6 \times 10^{5} P a; V_{A} = 0.20 l i t r e$
$V_{D} = 1.30 l i t r e$
Find the work performed (in joules) by the system along path $A D$ .