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Question

Physics Question on Thermodynamics

A quantity of a substance in a closed system is made to undergo a reversible process from an initial volume of 3m33 \,m^3 and initial pressure 105N/m210^5\, N/m^2 to a final volume of 5m35\, m^3 . If the pressure is proportional to the square of the volume (i.e. p=AV2p = AV^2 ), the work done by the substance will be

A

3.6×102J3.6 \times 10^2\,J

B

7.4×103J7.4 \times 10^3\,J

C

2.2×104J2.2 \times 10^4\, J

D

3.6×105J3.6 \times 10^5\,J

Answer

3.6×105J3.6 \times 10^5\,J

Explanation

Solution

Given, p=AV2p =AV^2
For a closed system, ΔQ=0\Delta Q = 0
ΔW=ΔU\therefore \Delta W = \Delta U
Also, work done by the substance
W=vivtpdVW = \int\limits_{v_i}^{v_{t}} p \cdot dV
=vivtAV2dV= \int\limits_{v_i}^{v_{t}} AV^{2} dV
=A(V33)vivt(A=piVi2)= A \left(\frac{V^{3}}{3}\right)_{v_i}^{v_{t}} \,\,\, \left(\because A = \frac{p_{i}}{V_{i}^{2}}\right)
where, AA is constant.
=A3[Vf3Vi3]= \frac{A}{3} \left[V_{f}^{3} -V_{i}^{3}\right]
=pi3Vi(Vf3Vi3)= \frac{p_{i}}{3V_{i}} \left(V_{f}^{3} - V_{i}^{3}\right)
=1053×9(12527)= \frac{10^{5}}{3\times 9} \left(125 - 27\right)
=1053×9×98= \frac{10^{5}}{3\times 9} \times 98
=3.6×105J= 3.6 \times 10^{5}\,J