Quasi-Satic Adiabatic Process

[shaiqbashir]

Hello Friends!

can anybody help me?

show that for a quasi static adiabatic process of an ideal gas, the following reaction holds

T$$V^{\gamma-1}}=constant$$

Secondly:::

At about 100ms after detonation of a uranium fission bomb, the "ball of fire" cosists of a sphere of gas with a radius of about 50 ft and a temperature of 300,000 degree Kelvin. At what radius is the temperature 2000 K?

 

[Rocketa]

For the First:

Use Ideal Gas Law and the Equation for adiabatic process $$PV^{\gamma}=constant$$

 

[ravenprp]

Sure, I can help you with that! A quasi-static adiabatic process is a thermodynamic process where the system changes very slowly, so that the system can be considered in equilibrium at all times. This means that the system is always close to its thermodynamic equilibrium state and there is no heat transfer between the system and its surroundings. In this type of process, the internal energy of the system remains constant, so the first law of thermodynamics reduces to:

\Delta U = W

where \Delta U is the change in internal energy and W is the work done on or by the system. For an ideal gas, the internal energy is a function of temperature only, so we can write:

\Delta U = C_V\Delta T

where C_V is the heat capacity at constant volume. Since there is no heat transfer, \Delta U = 0, and we can rearrange the equation to get:

C_V\Delta T = W

Now, for an adiabatic process, the work done can be expressed as:

W = -P\Delta V

where P is the pressure and \Delta V is the change in volume. For an ideal gas, we also know that the pressure and volume are related by the ideal gas law:

PV = nRT

where n is the number of moles, R is the gas constant, and T is the temperature. So, we can substitute this into our equation for work to get:

W = -nRT\frac{\Delta V}{V}

Now, for an adiabatic process, the change in volume is related to the change in temperature by the equation:

\frac{\Delta V}{V} = \gamma\frac{\Delta T}{T}

where \gamma is the adiabatic index, which for an ideal gas is equal to the ratio of specific heat capacities, \gamma = \frac{C_P}{C_V}. Substituting this into our equation for work, we get:

W = -nRT\gamma\frac{\Delta T}{T}

Finally, we can equate this to our equation for internal energy and solve for \Delta T:

C_V\Delta T = -nRT\gamma\frac{\Delta T}{T}

\Delta T\left(C_V + nR\gamma\right) = 0

Since \Delta T cannot be zero, we can divide both sides by \Delta T and rearrange to