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julsineya [31]
3 years ago
15

A piston-cylinder assembly contains air at a pressure of 30 lbf/in2 and a volume of 0.75 ft3. The air is heated at constant pres

sure until its volume is doubled. Assuming the ideal gas model, with constant specific heat ratio of k=1.4 for the air, determine the work and heat transfer, each in BTU.
Physics
1 answer:
garri49 [273]3 years ago
8 0

Answer:

Work W=4.16 Btu

Heat transfer Q=14.56 Btu

Explanation:

To calculate the work W we will use the given pressure p=30 psia and volume V₁=0.75 ft³ V₂=2.V₁

W=p.(V_{2}-V_{1} )\\W=30psia.(1.5ft^{3}-0.175ft^{3} ).\frac{144in^{2} }{1ft^{2} } .\frac{1Btu}{778lbf.ft}\\ W=4.16Btu

Now for heat transfer

Q=W+m.(u_{2}-u_{1} )\\Q=W+\frac{m.R.(T_{2}- T_{1})}{k-1}\\

The last equation we will rewrite using ideal gas law to calculate heat transfer

Q=W+\frac{p(V_{2}-V_{1} )}{k-1}\\ As\\W=p(V_{2}-V_{1} )\\So\\Q=W+\frac{W}{k-1} \\Q=4.16Btu+\frac{4.16Btu}{1.4-1} \\Q=14.56Btu

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- Attachment for figures missing in the question.

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Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

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Part a)

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- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

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                                  r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

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Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

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