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

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:

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

W = amount of work done = ?

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

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

$V_2$ = final volume = $(2\times V_1)=(2\times 6000)=12000L$

Putting values in above equation, we get:

$W=-1.28atm\times (12000-6000)L=-7680L.atm$

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

$1L.atm=101.33J$

1 kJ = 1000 J

So, $-7680L.atm=-35\times 101.3=-777984J=-778kJ$

The negative sign indicates the system is doing work.

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

Answer 2

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)