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asambeis [7]
3 years ago
5

20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu

me of the gas? (Assume that there's no change in temperature.)
A. 100 cubic inches
B. 10 cubic inches
C. 40 cubic inches
D. 5 cubic inches

The correct answer is B) 10 cubic inches
Physics
2 answers:
boyakko [2]3 years ago
8 0

It is given that that initial volume of gas is 20 cubic inches.

The initial pressure is given as 5 psi.

The final pressure is given as 10 psi

We are asked to calculate the final volume.

Let initial and final volume of gas is denoted as v_{1} \ and\ v_{2}

Let the initial and final pressure of gas is denoted as p_{1} \ and\ p_{2}

As per the question the temperature of a gas is constant.

From Boyle's law we know that pressure of a given mass of a gas is inversely proportional to the applied volume at constant temperature.

Hence mathematically

                                p\ \alpha\ \frac{1}{v}

                                pv=constant

                                p_{1} v_{1} =p_{2} v_{2}

                                v_{2} =\frac{p_{1}v_{1} }{p_{2} }

Putting the values of these respective quantities as mentioned above we get-

                                  v_{2} =\frac{5\ psi*20 cubic\ inches}{10\ psi}

                                         =10\ cubic\ inches        [ans]    


Mekhanik [1.2K]3 years ago
4 0
PoVo = PfVf    Vf=PoVo/Pf  = 5*20/10 = 100/10 = 10 cubic inches
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one mole of water is equivalent to 18 grams of water. a glass of water has a mass of 200 g. how many moles of water is in this?
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3 years ago
An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and
Andre45 [30]

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

v_1 = vf + x(vg - vf)

      = 0.001053 + 0.25(1.1594 - 0.001053)

v_1 = 0.20964  m^3/kg

s_1 = Sf + x(Sfg)

     = 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume  sov_1 -  v_2  = 0.29064 m^3/kg

v_2 = v_1 = vf + x(vg - vf)

       =0.29064 = x_2(0.88578 - 0.001061)

x_2 = 0.327

s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K

Change in entropy \Delta s = m(s_2 - s_1)

              =3( 3.36035 - 2.88) =  1.44 kJ/kg

8 0
3 years ago
A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.905 m, as the drawing shows. At the top of
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Answer:

5.959 m/s

Explanation:

m = Mass of gymnast

u = Initial velocity

v = Final velocity

h_i = Initial height

h_f = Final height

From conservation of Energy

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h_i=2r

v=\sqrt{4gr}\\\Rightarrow v=\sqrt{4\times 9.81\times 0.905}\\\Rightarrow v=5.959\ m/s

Velocity of gymnast at bottom of swing is 5.959 m/s

5 0
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