Ask question

Ask question

All categories

3 years ago

12

The density of oxygen contained in a tank is 2.0 kg/m3 when the temperature is 25 °C. Determine the gage pressure of the gas if

the atmospheric pressure is 97 kPa.

1 answer:

Verdich [7]3 years ago

8 0

Answer:

57.8408 kPa

Explanation:

Data provided in the question:

Density of oxygen contained, ρ = 2.0 kg/m³

Temperature, T = 25 °C = 25 +273 = 298 K

Atmospheric pressure = 97 kPa.

Now,

Pressure due to gas = ρRT

Here,

R is the gas constant = 259.8 J/kg.K

Thus,

Absolute Pressure due to oxygen = 2 × 259.8 × 298

= 154840.8 Pa

= 154840.8 × 10⁻³ kPa

= 154.84 kPa

Thus,

Gage pressure = Absolute Pressure - Atmospheric pressure

= 154.84 kPa - 97 kPa

= 57.8408 kPa

You might be interested in

Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi

qaws [65]

Answer:

$\Delta V = 209.151\,L$ , $\Delta C = 217.517\,USD$

Explanation:

The drag force is equal to:

$F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A$

Where $C_{D}$ is the drag coefficient and $A$ is the frontal area, respectively. The work loss due to drag forces is:

$W = F_{D}\cdot \Delta s$

The reduction on amount of fuel is associated with the reduction in work loss:

$\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s$

Where $F_{D,1}$ and $F_{D,2}$ are the original and the reduced frontal areas, respectively.

$\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s$

The change is work loss in a year is:

$\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)$

$\Delta W = 2.043\times 10^{9}\,J$

$\Delta W = 2.043\times 10^{6}\,kJ$

The change in chemical energy from gasoline is:

$\Delta E = \frac{\Delta W}{\eta}$

$\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}$

$\Delta E = 6.81\times 10^{6}\,kJ$

The changes in gasoline consumption is:

$\Delta m = \frac{\Delta E}{L_{c}}$

$\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }$

$\Delta m = 154.772\,kg$

$\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }$

$\Delta V = 209.151\,L$

Lastly, the money saved is:

$\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )$

$\Delta C = 217.517\,USD$

4 0

3 years ago

Which type of boot authentication is more secure?

MA_775_DIABLO [31]

The type of boot authentication that is more secure is Unified Extensible Firmware Interface

Unified Extensible Firmware Interface help to provide a computer booting that is more secured.

Unified Extensible Firmware Interface is a computer software program that work hand in hand with an operating system, it main function is to stop a computer system from boot with an operating system that is not secured.

For a computer system to boot successfully it means that the Operating system support the Unified Extensible Firmware Interface because it secured.

Inconclusion The type of boot authentication that is more secure is Unified Extensible Firmware Interface

Learn more here :

brainly.com/question/24750986

7 0

2 years ago

An example of the split-off point in oil, gasoline, and kerosene production is that point where crude oil is

eimsori [14]

i believe the correct answer is c but i’m sorry if i’m not correct

8 0

3 years ago

Find the Thevenin Equivalent.

arlik [135]

Answer:

hindi ko po gets

Explanation:

paulit po ng picture

4 0

3 years ago

Read 2 more answers

For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _______ and using

serious [3.7K]

Answer:

Option A - fail/ not fail

Explanation:

For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _fail______ and using the Maximum Shear Stress Theory the material will ___not fail_______

6 0

3 years ago