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Blizzard [7]
3 years ago
11

Joules are Nm, which is kg middot m^2/s^2 Use this relationship and 1 Btu = 1.0551 kJ to determine what Btu using primary Englis

h units (ft, lb_m, degree F, sec, etc.)
Physics
1 answer:
son4ous [18]3 years ago
3 0

Answer:

23851.35840\ lbft^2/s^2

Explanation:

1\ Btu=1.0051\ kJ\\\Rightarrow 1\ Btu=1.0051\times 1000 kgm^2/s^2

The English units that are to be used here are pounds (lb) and feet (ft)

1\ kg=2.20462\ lb

1\ m=3.28084\ ft

1005.1\times 2.20462\times 3.28084^2=23851.35840\ lbft^2/s^2

The answer is 1\ Btu=23851.35840\ lbft^2/s^2

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A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then
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Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

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Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

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Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J

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720/180=4

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