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guapka [62]
3 years ago
9

Kim has a metal casting company which makes commemorative coins. She has 0.12 cubic meters of silver which she needs to make int

o 1.07e5 coins of thickness 2 mm. If she uses all of the silver, what is the diameter in centimeters of each coin
Physics
1 answer:
a_sh-v [17]3 years ago
5 0

Answer:

The coin has a diameter of 2.67 cm

Explanation:

First, we need to find the volume of each coin by dividing the total volume of silver by the number of coins. We have also to do a conversion of units in terms of centimeters as follows:

V=0.12 m^3\times (\frac{100cm}{1m})^3=12000\ cm^3\\V_c=\frac{120000\ cm^3}{1.07\times10^5}= 1.121 cm^3

Then, we define the coin as a tiny cilinder to determine its diameter. In that order we use the cilinder's volumen equation as follows:

V=\pi r^2h\\r = \sqrt \frac{V}{\pi h}= \sqrt\frac{1.121 cm^3}{\pi \times 0.2cm}=1.336 cm

Finally, we know that the diameter is twice the radius, therefore the diameter of each coin is 2.67 cm.

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        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

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               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

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          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

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                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

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