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nordsb [41]
3 years ago
13

700.0 liters of a gas are prepared at 760.0 mmHg and 100.0 °C. The gas is placed into a tank under high pressure. When the tank

cools to 32.0 °C, the pressure of the gas is 20.0 atm. What is the volume of the gas?
Engineering
1 answer:
ololo11 [35]3 years ago
5 0

Answer:

The volume of the gas is 11.2 L.

Explanation:

Initially, we have:

V₁ = 700.0 L

P₁ = 760.0 mmHg = 1 atm

T₁ = 100.0 °C

When the gas is in the thank we have:

V₂ =?

P₂ = 20.0 atm

T₂ = 32.0 °C      

Now, we can find the volume of the gas in the thank by using the Ideal Gas Law:

PV = nRT

V_{2} = \frac{nRT_{2}}{P_{2}}    (1)

Where R is the gas constant

With the initials conditions we can find the number of moles:

n = \frac{P_{1}V_{1}}{RT_{1}}    (2)

By entering equation (2) into (1) we have:

V_{2} = \frac{P_{1}V_{1}}{RT_{1}}*\frac{RT_{2}}{P_{2}} = \frac{1 atm*700.0 L*32.0 ^{\circ}}{100.0 ^{\circ}*20.0 atm} = 11.2 L

Therefore, When the gas is placed into a tank the volume of the gas is 11.2 L.

I hope it helps you!                                                                                                                                                                                

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Readme [11.4K]

Answer:Yes,266.66 MPa

Explanation:

Given

Yield stress of material =140 MPa

Cross-section of 300\times 100 mm^2

Force(F)=8 MN

Therefore stress due to this Force(\sigma)

\sigma =\frac{F}{A}=\frac{8\times 10^6}{300\times 100\times 10^{-6}}

\sigma =266.66 \times 10^{6} Pa

\sigma =266.66 MPa

Since induced stress  is greater than Yield stress therefore Plastic deformation occurs

8 0
3 years ago
Underground water is to be pumped by a 78% efficient 5- kW submerged pump to a pool whose free surface is 30 m above the undergr
maksim [4K]

Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

\eta_t = The efficiency of the pump = 78%

\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}

Q_{max} = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump Q_{max} ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

6 0
2 years ago
A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round
Harlamova29_29 [7]

Answer:

\frac{Q_{square}}{Q_{circle}} =  0.785  

Explanation:

given data

types of drinking straws

  1. square cross-sectional shape
  2. round shape

solution

we know that both perimeter of the cross section are equal

so we can say that

perimeter of square  = perimeter of circle  

4 × S = π × D

here S is length and D is diameter

S = \frac{\pi D}{4}        ....................1

and

ratio of  flow rate through the square and circle is here

\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}  

\frac{Q_{square}}{Q_{circle}} = \frac{S^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{(\frac{\pi D}{4})^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{\pi }{4}  

\frac{Q_{square}}{Q_{circle}} =  0.785  

4 0
3 years ago
Traffic at a roundabout moves
sweet-ann [11.9K]

Answer:

b-counter-clockwise

Explanation:

3 0
3 years ago
A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per
mr_godi [17]

Answer:

-50.005 KJ

Explanation:

Mass flow rate = 0.147 KJ per kg

mass= 10 kg

Δh= 50 m

Δv= 15 m/s

W= 10×0.147= 1.47 KJ

Δu= -5 kJ/kg

ΔKE + ΔPE+ ΔU= Q-W

0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W

Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu

= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50

= 1.47 +3.375-4.8450-50

Q=-50.005 KJ

7 0
3 years ago
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