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3 years ago

This equation is used to calculate the properties of a gas under nonideal conditions.

Physics

1 answer:

goblinko [34]3 years ago

5 0

Answer:

Van der Waal's equation

Explanation:

The Van der Waal's equation is use to calculate the properties of a gas under nonideal or real gases conditions.

$(P + \frac{an^2}{V^2})(V-nb) = nRT$ .

Here P, V ,T ,n and R have usual meaning as in the ideal gas equation

that is PV=nRT

with the difference of constant a and b. a and b are constants representing magnitude of intermolecular attraction and excluded volume respectively respectively.

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Q 24, 25, 26 i dont get them and need answers for it

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Answer:

24) W = 75 [J]; 25) W = 1794[J]; 26) n = 8.8 (times) or 9 (times)

Explanation:

24) This problem can be solved by means of the following equation.

$DU = Q-W\\$

where:

DU = internal energy difference [J]

Q = Heat transfer [J]

W = work [J]

Since there are no temperature changes the internal energy change is equal to zero

DU = 0

therefore:

$Q = W\\$

The work is equal to the heat transfered, W = 75 [J].

25) The heat transfer can be calculated by means of the following equation.

$Q = m*c_{p}*DT\\where:\\m = mass = 0.4[kg]\\c_{p} = specific heat = 897[J/kg*K]\\DT= 5 [C]$

Q = 0.4*897*5 = 1794[J]

Work is equal to heat transfer, W = 1794[J]

26) Each time the bag falls the potential energy is transformed into heat energy, which is released into the environment. In this way the potential energy is equal to the developed heat.

$E_{p}=Q\\\\E_{p}=m*g*h$

where:

m = mass = 0.5[kg]

g = gravity = 9.81[m/s^2]

h = 1.5 [m]

$E_{p}=0.5*9.81*1.5\\E_{p}=7.36[J]$

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$Q=m*c_{p}*DT\\Q=0.5*130*1\\Q=65[J]$

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At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

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Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

$K = - V\cdot \frac{dP}{dV}$

Where:

$K$ - Bulk module, measured in pascals.

$V$ - Sample volume, measured in cubic meters.

$P$ - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

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This resultant expression is solved by definite integration and algebraic handling:

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$\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }$

The final volume is predicted by:

$V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }$

If $V_{o} = 1\,m^{3}$ , $P_{o} - P_{f} = -10132500\,Pa$ and $K = 2.3\times 10^{9}\,Pa$ , then:

$V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }$

$V_{f} \approx 0.996\,m^{3}$

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$\Delta V = V_{o} - V_{f}$

$\Delta V = 1\,m^{3} - 0.996\,m^{3}$

$\Delta V = 0.004\,m^{3}$

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