BE SURE ANSWER IS IN THE CORRECT UNITS! THANK YOU!! Consider the following figure. (The x axis is marked in increments of 2.5 m³.)P (Pa)6 × 1064 × 1062 × 106V (m³)(a) Determine the work done on a gas that expands from i to f as indicated in the figure.The work done by the gas will be positive when it expands. The problem asks for the work done on the gas. MJ(b) How much work is performed on the gas if it is compressed from f to i along the same path?The work done by the gas will be negative when it is compressed. The problem asks for the work done on the gas. MJNeed Help?Read ItWatch It

Answered: Consider the following figure. (The x…

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Chapter1: Units, Trigonometry. And Vectors

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BE SURE ANSWER IS IN THE CORRECT UNITS! THANK YOU!!

Consider the following figure. (The x axis is marked in increments of 2.5 m³.)
P (Pa)
6 × 106
4 × 106
2 × 106
V (m³)
(a) Determine the work done on a gas that expands from i to f as indicated in the figure.
The work done by the gas will be positive when it expands. The problem asks for the work done on the gas. MJ
(b) How much work is performed on the gas if it is compressed from f to i along the same path?
The work done by the gas will be negative when it is compressed. The problem asks for the work done on the gas. MJ
Need Help?
Read It
Watch It

Expert Solution

Step 1

Work done = Area under the curve in (P-V) coordinates .

In P-V diagram ,if it is a left to right process ,it indicates work is done by the gas. (Expansion process)

In P-V diagram ,if it is a right to left process ,it indicates work is done on the gas. (Compression Process)

Work done by the gas in expanding gas from i to f is found by the area under the graph.

${(W_{b y g a s})}_{i \to f} = (6 \times 10^{6} \times 2.5) + (\frac{(6 + 2) 10^{6}}{2} \times 2.5) + (2 \times 10^{6} \times 2.5) J {(W_{b y g a s})}_{i \to f} = (15 + 10 + 5) \times 10^{6} J {(W_{b y g a s})}_{i \to f} = 30 \times 10^{6} J {(W_{b y g a s})}_{i \to f} = 30 MJ$

Therefore, work done by the gas is 30 MJ.

Now, we know that

${(W_{o n g a s})}_{i \to f} = - {(W_{b y g a s})}_{i \to f} {(W_{on gas})}_{i \to f} = - 30 MJ$

Work done on the gas from i to f is -30 MJ.

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