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Tanya [424]
3 years ago
5

A closed system consists of 0.3 kmol of octane occupying a volume of 5 m³. Determine (a) the weight of the system, in N, and (b)

the molar- and mass-based specific volumes, in m³/kmol and m³/kg respectively. Let g=9.81m/s².
Engineering
1 answer:
Leni [432]3 years ago
7 0

Answer:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

Explanation:

To calculate the mass of the octane(m):

Number of mole of octane (n) =0.3kmol(given)

Molarmass of octane (M) =114.23kg/kmol

m=n*M

m=(0.3kmol)*(114.23kg/kmol)

m=34.269kg

To calculate for the weight of octane(W):

W=g*m

W=(9.81m/s^2)*(34.269kg)

W=336.18N

b) For specific volumes of Vn and Vm:

Given volume of octane (V) =5m^3

Vm=V/m

Vm=5m^3/34.269kg

Vm=0.1459m^3/kg

And Vn will be :

Vn=V/m=5m^3/0.3kmol

Vn=16.67m/Kmol

Therefore, the answers are:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

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3 years ago
The 10-lb block is pressed against the spring so as to compress it 2 ft when it is at a point A. If the plane is smooth, determi
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Answer:

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Substituting, we have

0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²

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V = √1094.796

V = 33.09 ft/s

But tan ∅ = 3/4

∅ = tan⁻¹3/4

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From uniform acceleration,

S = S₀ + ut + 1/2gt²

It can be written as

S = S₀ + Vsin∅*t + 1/2gt²

Substituting, we have

0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)

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Substituting, we have

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