Question

An ideal gas goes through the following two-step process. 1) The container holding the gas has a fixed volume of 0.200 m3 while the pressure of the gas increases from 3.00×105 Pa to 4.00×105 Pa . 2) The container holding the gas is then compressed to a volume of 0.120 m3 while maintaining a constant pressure of 4.00×105 Pa .A) What is the total work done by the gas for this two-step process?

Answer 1

`W=-3.2* 10^4\ J`

Step-by-step explanation:

Given:

Process 1:

• Volume of ideal gas is constant,
  `V_1_i=0.2\ m^3`

• Initial pressure,
  `P_1_i=3* 10^5\ Pa`

• Final pressure,
  `P_1_f=4* 10^5\ Pa`

Process 2:

• Pressure of ideal gas is constant,
  `P_2_f=4* 10^5\ Pa`

• Final volume,
  `V_2_f=0.12\ m^3`

We know that the work done by an ideal gas is given as:

`W=P* (V_f-V_i)`

Now for process 1:

`W_1=0\ J`

∵there is no change in volume in this process.

For process 2:

`W_2=4* 10^5\time (0.12-0.2)\ J`

`W_2=-3.2* 10^4\ J`

∵Negative , sign indicates that the work is being done on the gas here since the gas is being compressed.

Hence the total work done by the gas during this two step process is :

`W=W_1+W_2`

`W=0-3.2* 10^4`

`W=-3.2* 10^4\ J` is the work done by the gas.