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Question: For the adiabatic expansion of an ideal gas : a.) \(P{V^\Upsilon }\) = constant b.) \(T{V^{\Upsi...

For the adiabatic expansion of an ideal gas :
a.) PVΥP{V^\Upsilon } = constant
b.) TVΥ−1T{V^{\Upsilon - 1}} = constant
c.) TP1−ΥT{P^{1 - \Upsilon }} = constant
d.) All of above

Explanation

Solution

The adiabatic expansion is the one which will not involve transfer of mass or heat between the system and its surroundings.
For an adiabatic expansion of an ideal gas,
PVΥP{V^\Upsilon } = constant
Where Υ=CPCV\Upsilon = \dfrac{{{C_P}}}{{{C_V}}}
The ideal gas equation is PV = nRT
Where n is the number of moles of gas.
For n=1 i.e. for one mole of gas,
PV = RT
Complete step by step answer:
The first thing we need to understand is the adiabatic process.
An Adiabatic process is the one in which there is no transfer of heat or mass between the thermodynamic system and surroundings. The energy is transferred in the form of work.
The expansion of a gas is directly related to temperature according to Charlie’s law. The gas expands with lowering in the temperature.
For the adiabatic expansion of a gas, the first law can be written as-
∂U=−PdV\partial U = - PdV
Let it be equation 1.
Further, we have CV=(∂U∂T)V{C_V} = {\left( {\dfrac{{\partial U}}{{\partial T}}} \right)_V}
Let it be equation 2.
As we have assumed ideal gas and we know in ideal gas there are no intermolecular forces. Thus, internal energy of gas will depend only on temperature. Volume has nothing to do with it.
So, we can write the above equation 2 as-
CV=(∂U∂T){C_V} = \left( {\dfrac{{\partial U}}{{\partial T}}} \right)
On rearranging, we can write- dU=CVdTdU = {C_V}dT
From equation 1, we have ∂U=−PdV\partial U = - PdV.
Thus, CVdT=−PdV{C_V}dT = - PdV
Let it be equation 3.
Further, the ideal gas equation is PV = nRT
Where n is number of moles of gas.
For n=1 i.e. one mole of gas,
PV = RT
Substituting it in equation 3, we have

PV = RT = ({C_P} - {C_V})T \\\ P = \dfrac{{({C_P} - {C_V})T}}{V} \\\ \end{gathered} $$ Let it be equation 4. From equation 3, we can write $$P = - \dfrac{{{C_V}dT}}{{{dV}}}$$ Substituting it in equation 4. $$- \dfrac{{{C_V}dT}}{{{dV}}} = \dfrac{{{C_P}-{C_V}T}}{{{V}}}$$ Rearranging it, we get ${C_V}dT = - ({C_P} - {C_V})\dfrac{{TdV}}{V}$ Let, $$\dfrac{{{C_P}}}{{{C_V}}} = \Upsilon $$ Now, we will separate the variables and fill the above value. We have, $$\dfrac{{dT}}{T} + (\Upsilon - 1)\dfrac{{dV}}{V}$$ = 0 On integrating the equation, $T{V^{\Upsilon - 1}}$ = constant Let it be equation 5. Thus, option b.) is the correct answer. Further, we have $\dfrac{{PV}}{T}$ = constant Let it be equation 6. On putting value of T from equation 5 in equation 6, we have $P{V^\Upsilon }$ = constant Thus, option a.) is the correct answer. We have ideal gas equation for one mole, PV = RT Substituting above, we can get $T{P^{1 - \Upsilon }}$ = constant Thus, option c.) is also the correct answer. **So, we can say that our answer will be option d.) i.e. All of the above.** **Note:** The ideal gas we talk about is not the real concept. It is an hypothetical concept. Ideal gas refers to the gas in which molecules occupy negligible space and zero forces of interactions. Further, the ideal gas obeys all the gas laws without a little deviation. A very good example of adiabatic process can be the flow of air vertically in the atmosphere. The air in our atmosphere rises up. As it rises, it expands and the temperature is lowered. The negative sign in equation 3 means that volume is increasing with decrease in temperature.