Acontainer 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

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:

W = amount of work done = ?

P = pressure of the container =    (Conversion factor:    )

= initial volume =     (Conversion factor:    )

= final volume =

Putting values in above equation, we get:

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

1 kJ = 1000 J

So,

The negative sign indicates the system is doing work.

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

it's all four options.