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Question: Calculate the work done for \( B \to A \) ![](https://www.vedantu.com/question-sets/45714ac3-518a...

Calculate the work done for B→AB \to A

(A) 6×10−3J6 \times {10^{ - 3}}J
(B) 12×10−3J12 \times {10^{ - 3}}J
(C) 3×10−3J3 \times {10^{ - 3}}J
(D) 4×10−3J4 \times {10^{ - 3}}J

Explanation

Solution

from B to A, observe that the pressure changes linearly. Generally work done can be defined as the integration of pressure with respect to volume through the path taken.

Formula used : In this solution we will be using the following formulae;
W=∫PdVW = \int {PdV} where WW is the work done on a gas, PP is the pressure of the gas and VV is the volume of the gas.

Complete step by step answer:
To calculate the work done, we note that the pressure is linearly changing with volume, hence, we can model it as an equation of a straight line as in
x=my+cx = my + c
Then, P=mV+cP = mV + c
Now, from the diagram, we have that at volume equal to 2 litres (which is 2×10−3m32 \times {10^{ - 3}}{m^3} ), pressure is 5N/m25N/{m^2} .
Hence, 5=m(2×10−3)+c5 = m\left( {2 \times {{10}^{ - 3}}} \right) + c
And similarly from diagram, we also get
1=m(4×10−3)+c1 = m\left( {4 \times {{10}^{ - 3}}} \right) + c
Solving the two equations simultaneously, we get that
m=−2×103m = - 2 \times {10^3} and c=9c = 9
Hence, P=−2×103V+9P = - 2 \times {10^3}V + 9
Now, the work done can be defined as
W=∫PdVW = \int {PdV} where WW is the work done on a gas, PP is the pressure of the gas and VV is the volume of the gas.
So by inserting expression and integrating, we have
W=∫VBVAPdV=∫2×10−34×10−3(−2×103V+9)dVW = \int_{{V_B}}^{{V_A}} {PdV} = \int_{2 \times {{10}^{ - 3}}}^{4 \times {{10}^{ - 3}}} {\left( { - 2 \times {{10}^3}V + 9} \right)dV}
Hence, W=[−103V2+9V]2×10−34×10−3W = \left[ { - {{10}^3}{V^2} + 9V} \right]_{2 \times {{10}^{ - 3}}}^{4 \times {{10}^{ - 3}}}
Hence, by carrying out the definite integral, we have that
W=−6×10−3JW = - 6 \times {10^{ - 3}}J
We can neglect the negative sign as it only means that work is done by the gas from B to A, hence, we get
W=6×10−3JW = 6 \times {10^{ - 3}}J
Hence, the correct option is A.

Note:
For clarity, we can write the equation of a straight line as x=my+cx = my + c . This is in contrast to the well known format of y=mx+cy = mx + c . However, it can be written the other way around too, which can be proven as in;
y=mx+cy = mx + c
x=y−cm=ym−cmx = \dfrac{{y - c}}{m} = \dfrac{y}{m} - \dfrac{c}{m}
But since mm and c are constants, we have that 1m\dfrac{1}{m} and cm\dfrac{c}{m} are constant. And there is the slope and intercept of the graph if xx is considered the dependent variable. Hence
x=m1y−c1x = {m_1}y - {c_1} .