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Question

Reversible adiabatic process may be expressed as (T1/T2) equal to

The correct answer is \({\left( {\frac{{{p_1}}}{{{p_2}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}\)

Understanding Reversible Adiabatic Processes

A reversible adiabatic process is a thermodynamic process where a system undergoes changes in its state variables (like pressure, volume, and temperature) without any heat exchange with its surroundings, and the process occurs in such a way that it can be reversed without leaving any trace on the surroundings. This means the entropy of the system remains constant throughout the process. For an ideal gas undergoing such a process, specific relationships exist between its state variables.

Adiabatic Process Relationships

For an ideal gas undergoing a reversible adiabatic process, the following relationships hold:

  • Between Pressure (P) and Volume (V): \(PV^\gamma = \text{constant}\)
  • Between Temperature (T) and Volume (V): \(TV^{\gamma-1} = \text{constant}\)
  • Between Temperature (T) and Pressure (P): \(T^{1-\gamma}P^\gamma = \text{constant}\) or equivalently \(T P^{-\frac{\gamma-1}{\gamma}} = \text{constant}\)

Here, \(\gamma\> is the adiabatic index, defined as the ratio of the specific heat at constant pressure (\(C_p\)) to the specific heat at constant volume (\(C_v\)), i.e., \(\gamma = \frac{C_p}{C_v}\).

Deriving Temperature-Pressure Relationship

We are interested in the relationship between the temperature ratio (\(\frac{T_1}{T_2}\)) and the pressure ratio (\(\frac{p_1}{p_2}\)). We can start from the relationship \(T^{1-\gamma}P^\gamma = \text{constant}\).

For two states 1 and 2 during the process:

\(T_1^{1-\gamma}P_1^\gamma = T_2^{1-\gamma}P_2^\gamma\)

To find the ratio \(\frac{T_1}{T_2}\), we rearrange the equation:

\(\frac{T_1^{1-\gamma}}{T_2^{1-\gamma}} = \frac{P_2^\gamma}{P_1^\gamma}\)

This can be written as:

\({\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^{1-\gamma}} = {\left( {\frac{{{p_2}}}{{{p_1}}}} \right)^\gamma}\)

Now, raise both sides to the power of \(\frac{1}{1-\gamma}\):

\(\frac{T_1}{T_2} = {\left( {\frac{{{p_2}}}{{{p_1}}}} \right)^{\frac{\gamma}{{1-\gamma}}}}\)

Using the property \({(a/b)}^{-x} = (b/a)^x\), we can rewrite the exponent:

\(\frac{\gamma}{{1-\gamma}} = \frac{-\gamma}{{\gamma-1}} = - \frac{\gamma}{{\gamma-1}}\)

So,

\(\frac{T_1}{T_2} = {\left( {\frac{{{p_1}}}{{{p_2}}}} \right)^{\frac{\gamma}{{\gamma-1}}}}\)

Alternatively, starting from \(T P^{-\frac{\gamma-1}{\gamma}} = \text{constant}\):

\(T_1 P_1^{-\frac{\gamma-1}{\gamma}} = T_2 P_2^{-\frac{\gamma-1}{\gamma}}\)

Rearranging for the temperature ratio:

\(\frac{T_1}{T_2} = \frac{P_2^{-\frac{\gamma-1}{\gamma}}}{P_1^{-\frac{\gamma-1}{\gamma}}}\)

Which simplifies to:

\(\frac{T_1}{T_2} = {\left( {\frac{{{p_1}}}{{{p_2}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}\)

Conclusion

The correct expression for the ratio of temperatures (\(\frac{T_1}{T_2}\)) in a reversible adiabatic process in terms of the pressure ratio (\(\frac{p_1}{p_2}\)) is:

\(\frac{T_1}{T_2} = {\left( {\frac{{{p_1}}}{{{p_2}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}\)

This matches option 3 provided in the question.

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Important Questions from Thermodynamic Relations

  1. Identify the statement, which best describes the second law of thermodynamics :
  2. Which of the following statement(s) is/are true for an isothermal process ?
    (a) There is no change in enthalpy
    (b) There is no change in internal energy
    (c) There is change in temperature
    Choose the correct code :
  3. The sum of the internal energy and product of pressure and volume is known as :
  4. The relation for specific heats is given by

  5. The property relation for enthalpy change, dh is:

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