A container of gas is at a pressure of 1.3 x 10^5 pa and a volume of 6.0 m^3. how much work is done by the gas if it expands at a constant pressure to twice its initial volume

For the answer to the question above, since the process is carried out at constant pressure,
work = P (V1-V2)
= 1.3 E5 (6-12) = -7.8 E5 N.m (in this case the work is made by the system, so it is negative if work made on the system then it should be positive)

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Аноним Аноним · Answer 2 · 2019-12-20T06:40:48+01:00

Answer: The work done for the given process is -778 kJ

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

[tex]W=-P\Delta V=-P(V_2-V_1)[/tex]

W = amount of work done = ?

P = pressure of the container = [tex]1.3\times 10^5Pa=1.28atm[/tex] (Conversion factor: [tex]1atm=1.013\times 10^5Pa[/tex] )

[tex]V_1[/tex] = initial volume = [tex]6.0m^3=6000L[/tex] (Conversion factor: [tex]1m^3=1000L[/tex] )

[tex]V_2[/tex] = final volume = [tex](2\times V_1)=(2\times 6000)=12000L[/tex]

Putting values in above equation, we get:

[tex]W=-1.28atm\times (12000-6000)L=-7680L.atm[/tex]

To convert this into joules, we use the conversion factor:

[tex]1L.atm=101.33J[/tex]

1 kJ = 1000 J

So, [tex]-7680L.atm=-35\times 101.3=-777984J=-778kJ[/tex]

The negative sign indicates the system is doing work.

Hence, the work done for the given process is -778 kJ