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A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3 and the pressure increases from 1.00×105 Pa to 4.00×105
Question
A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3 and the pressure increases from 1.00×105 Pa to 4.00×105 Pa. The second process is a compression to a volume of 5.00×10−2 m3 at a constant pressure of 4.00×105 Pa. Find the total work done by the gas during both processes.
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Physics
5 years
2021-08-31T10:23:04+00:00
2021-08-31T10:23:04+00:00 1 Answers
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Answers ( )
The work done = – 2 x 10⁴ J
Explanation:
In the first case , the volume is kept constant and pressure varies .
In isothermal process , the work done
W₁ = V x ΔP
here V is the volume of gas and ΔP is the change in pressure
Thus W₁ = 0
Because there is no change in volume , therefore displacement is zero .
In second case pressure is constant , but volume changes
Thus W₂ = P x ΔV
here P is the pressure and ΔV is the change in volume
Therefore W₂ = 4 x 10⁵ x 5 x 10⁻² = 2 x 10⁴ J
The total work done W = – 2 x 10⁴ J
Because the work done in compression is negative .