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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?

Respuesta :

Answer:

[tex]W=-3.2\times 10^4\ J[/tex]

Explanation:

Given:

Process 1:

  • Volume of ideal gas is constant, [tex]V_1_i=0.2\ m^3[/tex]
  • Initial pressure, [tex]P_1_i=3\times 10^5\ Pa[/tex]
  • Final pressure, [tex]P_1_f=4\times 10^5\ Pa[/tex]

Process 2:

  • Pressure of ideal gas is constant, [tex]P_2_f=4\times 10^5\ Pa[/tex]
  • Final volume, [tex]V_2_f=0.12\ m^3[/tex]

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

[tex]W=P\times (V_f-V_i)[/tex]

Now for process 1:

[tex]W_1=0\ J[/tex]

∵there is no change in volume in this process.

For process 2:

[tex]W_2=4\times 10^5\time (0.12-0.2)\ J[/tex]

[tex]W_2=-3.2\times 10^4\ J[/tex]

∵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 :

[tex]W=W_1+W_2[/tex]

[tex]W=0-3.2\times 10^4[/tex]

[tex]W=-3.2\times 10^4\ J[/tex] is the work done by the gas.

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