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julsineya [31]
3 years ago
6

A fixed mass of gas has a volume of 25 cm³. The pressure of the gas is 100 kPa.

Physics
1 answer:
tekilochka [14]3 years ago
7 0

Answer:

67 kPa

Explanation:

Given that,

Initial volume, V₁ = 25 cm³

Initial pressure, P₁ = 100 kPa

Final volume, V₂ = 15 cm³

We need to find the change in pressure of the gas. The relation between the volume and pressure of a gas is given by :

P\propto \dfrac{1}{V}\\\\\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\\P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{100\times 25}{15}\\\\=166.66\ kPa

or

= 167 kPa

The change in pressure,

= P₂ - P₁

= 167 kPa - 100 kPa

= 67 kPa

Hence, the correct option is (a).

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wo parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area o
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Two parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10.

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Answer:

a

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   ii    Q_2 =  4.4604 *10^{-11} \ C

b

  Q_{eq} = 6.5844 *10^{-11} \ C

Explanation:

From the question we are told that

   The  voltage of the battery is  V  = 12.0  \ V

    The  plate area of each capacitor is  A  =  5.30 \ cm^2  =  5.30 *10^{-4} \ m^2

    The  separation between the plates is  d =  2.65 \ mm =  2.65 *10^{-3} \ m

     The permittivity of free space  has a value  \epsilon_o  =  8.85 *10^{-12} \  F/m

     The  dielectric constant of the other material is  z =  2.10

The  capacitance of the  first capacitor is mathematically represented as

       C_1  =  \frac{\epsilon  *  A }{d }

substituting values

        C_1  =  \frac{8.85 *10^{-12 } *   5.30 *10^{-4} }{2.65 *10^{-3} }

       C_1  =  1.77 *10^{-12} \  F

The  charge stored in the first capacitor is  

       Q_1 =  C_1 *  V

substituting values

        Q_1 =  1.77 *10^{-12} * 12

       Q_1 =  2.124 *10^{-11} \  C

The capacitance of the second  capacitor is mathematically represented as

       C_2  =  \frac{ z * \epsilon  *  A }{d }

substituting values

       C_1  =  \frac{  2.10 *8.85 *10^{-12 } *   5.30 *10^{-4} }{2.65 *10^{-3} }

       C_1  =  3.717 *10^{-12}  \ F

The  charge stored in the second capacitor is  

      Q_2 =  C_2 *  V

substituting values

     Q_2 = 3.717*10^{-12} *  12

     Q_2 =  4.4604 *10^{-11} \ C

Now  the total charge stored in the parallel combination is mathematically represented as

     Q_{eq} =  Q_1 + Q_2

substituting values

    Q_{eq} =  4.4604 *10^{-11} + 2.124*10^{-11}

     Q_{eq} = 6.5844 *10^{-11} \ C

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