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kupik [55]
3 years ago
13

What is the counter interval of this map?

Chemistry
1 answer:
lyudmila [28]3 years ago
3 0

Answer:

its the middle one "20"

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How much heat does it take to increase the temperature of 2.70 mol of an ideal gas by 30.0 K near room temperature if the gas is
Mnenie [13.5K]

Answer:

1683.6J

Explanation:

Given:

n= no. Of mol= 2.70 mol

T= Temperature= 30.0 K

Q= n Cv × ∆T .........eqn(1)

Where CV= molar heat capacity=5/2R for diatomic particle ,such as H2

CV= molar heat capacity=3/2R for diatomic, such as H

R= gas constant= 8.314 J/mol.K

Q= heat energy

For a diatomic molecules

Q= n Cv × T

But

Cv= molar heat capacity=5/2R = 5/2(8.314)=20.785

CV= 20.785

. ∆T= Temperature= 30.0 K

Then substitute the values into the eqn(1)

Q= 2.70 × 5/2(8.314) × 30

Q= 2.70 × 20.785 × 30

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The coefficient of thermal expansion α = (1/V)(∂V/∂T)p. Using the equation of state, compute the value of α for an ideal gas. Th
andreyandreev [35.5K]

Answer:

The coefficient of thermal expansion α is  

      \alpha  =  \frac{1}{T}

The coefficient of compressibility

      \beta   =  \frac{1}{P}

Now  considering (\frac{ \delta P }{\delta  T} )V

From equation (1) we have that

       \frac{ \delta P}{\delta  T}  =  \frac{n R }{V}

From  ideal equation

         nR  =  \frac{PV}{T}

So

     \frac{\delta P}{\delta  T}  =  \frac{PV}{TV}

=>  \frac{\delta  P}{\delta  T}  =  \frac{P}{T}

=>   \frac{\delta  P}{\delta  T}  =  \frac{\alpha }{\beta}

Explanation:

From the question we are told that

   The  coefficient of thermal expansion is \alpha  =  \frac{1}{V} *  (\frac{\delta V}{ \delta  P})  P

    The coefficient of compressibility is \beta  =  - (\frac{1}{V} ) *  (\frac{\delta V}{ \delta P} ) T

Generally the ideal gas is  mathematically represented as

        PV  =  nRT

=>      V  =  \frac{nRT}{P}  --- (1)

differentiating both side with respect to T at constant P

       \frac{\delta V}{\delta T }  =  \frac{ n R }{P}

substituting the equation above into \alpha

       \alpha  =  \frac{1}{V} *  ( \frac{ n R }{P})  P

        \alpha  = \frac{nR}{PV}

Recall from ideal gas equation  T =  \frac{PV}{nR}

So

          \alpha  =  \frac{1}{T}

Now differentiate equation (1) above with respect to  P  at constant T

          \frac{\delta  V}{ \delta P}  =  -\frac{nRT}{P^2}

substituting the above  equation into equation of \beta

        \beta  =  - (\frac{1}{V} ) *  (-\frac{nRT}{P^2} ) T

        \beta =\frac{ (\frac{n RT}{PV} )}{P}

Recall from ideal gas equation that

       \frac{PV}{nRT}  =  1

So

       \beta   =  \frac{1}{P}

Now  considering (\frac{ \delta P }{\delta  T} )V

From equation (1) we have that

       \frac{ \delta P}{\delta  T}  =  \frac{n R }{V}

From  ideal equation

         nR  =  \frac{PV}{T}

So

     \frac{\delta P}{\delta  T}  =  \frac{PV}{TV}

=>  \frac{\delta  P}{\delta  T}  =  \frac{P}{T}

=>   \frac{\delta  P}{\delta  T}  =  \frac{\alpha }{\beta}

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Answer:

Some of the orbits are nearly circular, while the moons farthest from Jupiter have more irregular orbits. The outer moon's orbit in the opposite direction in which Jupiter spins, which is unusual and indicates the moons were asteroids that were sucked into Jupiter's orbit after the initial system was formed.

Explanation:

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