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Alex
2 years ago
3

The specific volume of 5 kg of water vapor at 1.5 MPa, 440°C is 0.2160 m3 /kg. Determine (a) the volume, in m3 , occupied by the

water vapor, (b) the amount of water vapor present, in gram moles, and (c) the number of molecules.
Physics
1 answer:
telo118 [61]2 years ago
8 0

Answer:

(a) Volume=1.08 m³

(b) η=273.3 moles

(c) n=1.646×10²⁶ molecules

Explanation:

Given data

Mass m=5 kg

Pressure P=1.5 MPa

Temperature t=440°C=(440+273)K=713 K

Specific volume v=0.2160 m³/kg

To find

(a) Volume Occupied

(b) Amount of water vapor grams per mole

(c) Number of molecules

Solution

Let the system is in equilibrium and vapor is ideal gas

For part (a) volume occupied

volume=m_{mass} *v_{specific-volume}\\ volume=(5kg)*(0.216m^{3}/kg )\\volume=1.08m^{3}

For part (b) amount of water vapor grams per mole

Apply ideal gas law

PV=n.R.T\\n=\frac{PV}{R.T}\\ n=\frac{(1.5*10^{6} )*(1.08)}{(8.314)(713)}\\ n=273.3 moles

For (c) number of molecules

n_{molecules}=n*N_{Avogadro-number}\\n_{molecules}=273.3*(6.02214*10^{23} )\\ n_{molecules}=1.646*10^{26}molecules

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Bill leaves his 60 W desk lamp on every day, including weekends, for eight hours. After one month (30 days), how much total ener
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' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.

That's all the physics we need to know to answer this question.
The rest is just arithmetic.

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Wait a minute !  Hold up !  Hee haw !  Whoa ! 
Excuse me.  That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's

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Answer:

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5 0
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