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Law Incorporation [45]
3 years ago
11

A company that produces footballs uses a proprietary mixture of ideal gases to inflate their footballs. If the temperature of 23

0 grams [g] of gas mixture in a 15-liter [L] tank is maintained at 465 degrees Rankine [°R] and the tank is pressurized to 135 pound-force per square inch [psi], what is the molecular weight of the gas mixture in units of grams per mole [g/mol]?
Engineering
1 answer:
Arte-miy333 [17]3 years ago
5 0

Answer:

35 g/mol

Explanation:

Convert to SI units.

465 R × 5/9 = 258.3 K

Assuming the pressure is absolute pressure:

135 psi × (101.3 kPa / 14.7 psi) = 930.3 kPa

Ideal gas law:

PV = nRT

(930.3 kPa) (15 L) = n (8.314 kPa L / mol / K) (258.3 K)

n = 6.50 mol

The molar mass is therefore:

230 g / 6.50 mol = 35.4 g/mol

Rounded to two significant figures, the molar mass is 35 g/mol.

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Answer with Explanation:

Part a)

The volume of water in the tank as a function of time is plotted in the below attached figure.

The vertical intercept of the graph is 46.

Part b)

The vertical intercept represents the volume of water that is initially present in the tank before draining begins.

Part c)

To find the time required to completely drain the tank we calculate the volume of the water in the tank to zero.

0=46-3.5t\\\\3.5=46\\\\\therefore t=\frac{46}{3.5}=13.143minutes

Part d)

The horizontal intercept represents the time it takes to empty the tank which as calculated above is 13.143 minutes.

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Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat
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Answer:

V = 0.30787 m³/s

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Explanation:

given data

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steadily =200 kPa

temperature = 20°C

velocity = 5 m/s

solution

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velocity = mass × v

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v2 = A2×v2

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