Question

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

Answer 1

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)

Answer 2

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

Step-by-step explanation:

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

`W=-P\Delta V=-P(V_2-V_1)`

W = amount of work done = ?

P = pressure of the container =
`1.3* 10^5Pa=1.28atm` (Conversion factor:
`1atm=1.013* 10^5Pa` )

`V_1` = initial volume =
`6.0m^3=6000L` (Conversion factor:
`1m^3=1000L` )

`V_2` = final volume =
`(2* V_1)=(2* 6000)=12000L`

Putting values in above equation, we get:

`W=-1.28atm* (12000-6000)L=-7680L.atm`

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

`1L.atm=101.33J`

1 kJ = 1000 J

So,
`-7680L.atm=-35* 101.3=-777984J=-778kJ`

The negative sign indicates the system is doing work.

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