A gas undergoes two processes. In the first, the volume remains constant at 0.170 m3 and the pressure increases from 1.50×105 Pa to 6.00×105 Pa . The second process is a compression to a volume of 0.130 m3 at a constant pressure of 6.00×105 Pa. Find the total work done by the gas during both processes.

Explanation:

The work (W) done by the gas can be calculated using the following equation:

Where:

p: is the pressure

[tex}V_{f}[/tex]: is the final volume

[tex}V_{i}[/tex]: is the initial volume

In the first process, the work done by the gas is:

Since the volume remains constant, the total work done by the gas is equal to zero.

In the second process, the work done by the gas is:

Now, the total work done by the gas during both processes is:

Therefore, the total work done by the gas during both processes is - 24 kJ.

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answer: i don't know the answer good luck

explanation:

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2