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The volume of a gas decreases from 6 to 2 liters under a constant pressure of one atmosphere. Is the work done by the gas or

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Answer #1

a) Work is done on the gas

b) Workdone, W = P*(V2 - V1)

= 1.013*10^5*(2 - 6)*10^-3

= -405.2 J

here negative sign indicates that work is done on the system.

c) delta_U = -140 J

now use first law of thermodynamics,

Q = W + delta_U

= -405.2 - 140

= -545.2 J

here negative sign indicates that the heat released by the system.

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