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san4es73 [151]
3 years ago
11

A vessel containing 200 ml hydrogen gas at pressure 500torr at temperature 10 degree C.Find its volume when pressure increased t

o 700 torr and temperature 20degree C.
Physics
1 answer:
Margaret [11]3 years ago
3 0

Answer:

When the pressure and the temperature are increased the volume is 285.7 ml.

Explanation:

We can find the new volume by using the Ideal Gas Law:

PV = nRT

Where:

P: is the pressure

V: is the volume

n: is the number of moles

R: is the gas constant

T: is the temperature  

Initially, when V₁ = 200 ml, P₁ = 500 torr and T₁ = 10 °C, we have:

nR = \frac{P_{1}V_{1}}{T_{1}}   (1)  

And finally, when P₂ = 700 torr and T₂ = 20 °C, we have:

nR = \frac{P_{2}V_{2}}{T_{2}}   (2)

By equating (1) with (2):

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}                

V_{2} = \frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}} = \frac{500 torr*200 ml*20 ^{\circ} C}{10 ^{\circ} C*700 torr} = 285.7 ml

Therefore, when the pressure and the temperature are increased the volume is 285.7 ml.                                                

I hope it helps you!                  

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Alpha emission is the process results in a change in mass number. Option B is correct.

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Hence option B is correct.

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3 years ago
During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s
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Answer:

part a : <em>The dry unit weight is 0.0616  </em>lb/in^3<em />

part b : <em>The void ratio is 0.77</em>

part c :  <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}

Here

  • \gamma_d is the dry unit weight which is to be calculated
  • γ is the bulk unit weight given as

                                              \gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3

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Substituting values

                                              \gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3

<em>The dry unit weight is 0.0616  </em>lb/in^3<em />

Part b

Void Ratio

The void ratio is given as

                                                e=\frac{G_s \gamma_w}{\gamma_d} -1

Here

  • e is the void ratio which is to be calculated
  • \gamma_d is the dry unit weight which is calculated in part a
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72

Substituting values

                                              e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

S=\frac{G w}{e}

Here

  • e is the void ratio which is calculated in part b
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

                                      S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}

Here

  • \gamma_{zav} is  the zero air unit weight which is to be calculated
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

                                      \gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\

Now as the volume is known, the the overall weight is given as

weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

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