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Brums [2.3K]
3 years ago
14

A gas is in a sealed container.Part ABy what factor does the gas temperature change if the volume is doubled and the pressure is

tripled?Part BBy what factor does the gas temperature change if the volume is halved and the pressure is tripled?
Physics
1 answer:
Goryan [66]3 years ago
4 0

A) The temperature increases by a factor 6

We can use the ideal gas equation:

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

We can also rewrite it as

\frac{pV}{T}=nR

The gas is in a sealed container - this means the amount of gas is fixed, so n is constant. Since R is constant too, the term on the right in the equation is constant. So we can rewrite the equation as:

\frac{p_1V_1}{T_1}=\frac{p_2 V_2}{T_2}

where in this problem we have:

V_2 = 2V_1 (the volume is doubled)

p_2 = 3 p_1 (the pressure is tripled)

re-arranging the equation, we find the change in temperature:

\frac{T_2}{T_1}=\frac{p_2 V_2}{p_1 V_1}=\frac{(3 p_1)(2 V_1)}{p_1 V_1}=6

so, the temperature increases by a factor 6.

B) The temperature increases by a factor 1.5.

We can use again the same equation:

\frac{p_1V_1}{T_1}=\frac{p_2 V_2}{T_2}

Where in this case:

V_2 = \frac{V_1}{2} (the volume is halved)

p_2 = 3 p_1 (the pressure is tripled)

So, we can find the change in temperature:

\frac{T_2}{T_1}=\frac{p_2 V_2}{p_1 V_1}=\frac{(3 p_1)(\frac{V_1}{2})}{p_1 V_1}=\frac{3}{2}=1.5

So, the temperature increases by 1.5 times.

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AnnZ [28]
First, calculate the work done: 54 x 10 = 540J.
Now calculate power: 540 / 6 = 90.
The answer is 90W.

Hope this helps.
7 0
3 years ago
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One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.​
Zepler [3.9K]

Answer:

The answer is "\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}"

Explanation:

Given:

\to v=1\ liter= 10^{-3} \ m^3\\\\\to  w= 9.6 \ N\\

calculation:

Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978

4 0
2 years ago
1. What do you notice or wonder about ultrasound imaging? (black and white pictures)
a_sh-v [17]

Answer:

the shape how it involve into a picture

Explanation:

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3 years ago
Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the c
Amanda [17]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is  v_A= 4m/s

Explanation:

From the question we are told that

    The length of the string is  L = 1m

     The initial speed of block A is u_A

     The final speed of block A is  v_A = \frac{1}{2}u_A

      The initial speed of block B is u_B = 0

      The mass of block A  is  m_A = 7kg  gh

      The mass of block B is  m_B  = 2 kg

According to the principle of conservation of momentum

       m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}

Since block B at initial is at rest

       m_A u_A  = m_Bv_B + m_A \frac{u_A}{2}

      m_A u_A  - m_A \frac{u_A}{2} = m_Bv_B

          m_A \frac{u_A}{2} = m_Bv_B

  making v_B the subject of the formula

             v_B =m_A \frac{u_A}{2 m_B}

Substituting values

               v_B =\frac{7 u_A}{4}  

This v__B is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity  of v__B' at  the top of the vertical circle  

 The angular centripetal acceleration  would be mathematically represented

                   a= \frac{v^2_{B}'}{L}

Note that  this acceleration would be toward the center of the circle

      Now the forces acting at the top of the circle can be represented mathematically as

         T + mg = m \frac{v^2_{B}'}{L}

    Where T is the tension on the string

  According to the law of energy conservation

The energy at  bottom of the vertical circle   =  The energy at the top of

                                                                                the vertical circle

   This can be mathematically represented as

                 \frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L

From above  

                (T + mg) L = m v^2_{B}'

Substitute this into above equation

             \frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L  + mg 2L  

             \frac{49 mv_A^2}{16}  = \frac{1}{2} (T + mg) L + mg 2L

          \frac{49 mv_A^2}{16}  = T + 5mgL

The  value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is  zero

        This is mathematically represented as

                      \frac{49 mv_A^2}{16}  = 5mgL

making  v_A the subject

            v_A = \sqrt{\frac{80mgL}{49m} }

substituting values

          v_A = \sqrt{\frac{80* 9.8 *1}{49} }

              v_A= 4m/s

     

6 0
3 years ago
The atmosphere of Jupiter is essentially made up of hydrogen, H2. For H2, the specific gas constant is 4157 J/(kg K). The accele
Alenkinab [10]

Answer:

h=17357.9m

Explanation:

The atmospheric pressure is just related to the weight of an arbitrary column of gas in the atmosphere above a given area. So, if you are higher in the atmosphere less gass will be over you, which means you are bearing less gas and the pressure is less.

To calculate this, you need to use the barometric formula:

P=P_0e^{-\frac{Mg}{RT}h}

Where R is the gas constant, M the molar mass of the gas, g the acceleration of gravity, T the temperature and h the height.

Furthermore, the specific gas constant is defined by:

R_{H_2}=\frac{R}{M}

Therefore yo can write the barometric formula as:

P=P_0e^{-\frac{g}{R_{H_2}T}h}

at the surface of the planet (h =0) the pressure is P_0[\tex]. The pressure at the height requested is half of that:[tex]P=\frac{P_0}{2}

applying to the previuos equation:

\frac{P_0}{2} =P_0e^{-\frac{g}{R_{H_2}T}h}

solving for h:

h=17357.9m

3 0
3 years ago
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