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In this problem you are to consider an adiabaticexpansion of an ideal diatomic gas, which means that the gas expands with no addition or subtraction of heat.Assume that the gas is initially at pressure p0, volume V0, and temperature T0. In addition, assume that the temperature of the gas is such that you can neglect vibrational degrees of freedom. Thus, the ratio of heat capacities is γ=Cp/CV=7/5.Note that, unless explicitly stated, the variable γ should not appear in your answers--if needed use the fact that γ=7/5 for an ideal diatomic gas.Part AFind an analytic expression for p(V), the pressure as a function of volume, during the adiabatic expansion.Express the pressure in terms of V and any or all of the given initial values p0, T0, and V0.Correctp(V) = p0(V0V)75Part BAt the end of the adiabatic expansion, the gas fills a new volume V1, where V1>V0. Find W, the work done by the gas on the container during the expansion.Express the work in terms of p0, V0, and V1. Your answer should not depend on temperature.Part CFind ΔU, the change of internal energy of the gas during the adiabatic expansion from volume V0 to volume V1.Express the change of internal energy in terms of p0, V0, and/or V1.Need Part B and C

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Answer:

A: P=Po(Vo/V)^y

B: W=5/2Po(Vo-Vo^7/5 Vi^-2/5)

C: ∆V= 5/2Po[Vo^7/5 Vi^-2/5 -Vo]

Explanation:

Attached is the solution

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An analytical expression for P(V) is,   [tex]P=\left ( \frac{V_{0} }{V} \right )^{7/5} P_{0[/tex].

Work done is, [tex]W=\frac{5}{2} P_{0}V_{0}^{7/5} }\left ( V_{0}^{-2/5} \right - V_{1}^{\frac{-2}{5} } } )[/tex].

Internal Energy Change is,   [tex]\bigtriangleup U= \frac{5}{2} P_{0} \frac{P}\left ( V_{0}^{7/5 }V_{1}^{-7/5} \right- V_{0} )[/tex].

Given:-

The ratio of heat capacity is  [tex]\dfrac{7}{5}[/tex]

Initial pressure=0

Initial Volume=0

Initial tempreture=0

Part A-

An analytic expression for p(V),

It is known for an adiabatic process that,

[tex]PV^{\gamma}[/tex]=constant

Pressure(p) at volume (V),

[tex]PV^{\gamma}\\[/tex]=[tex]P_{0} V_{0}^{\gamma}[/tex]

[tex]P=\left ( \frac{V_{0} }{V} \right )^{\gamma} P_{0[/tex]

at γ =7/5

[tex]P=\left ( \frac{V_{0} }{V} \right )^{7/5} P_{0[/tex]

This is the required expression for P(V)

Part B-

Work done, the formula for work done can be written as,

[tex]W=\int\limits^{V_{1}} _{V_ {0}} P dv[/tex]

[tex]W=P_{0}V_{0}^{7/5} } \int\limits^{V_{1}} _{V_ {0}} \frac{1}{V^{7/5} } dv[/tex]

[tex]W=P_{0}V_{0}^{7/5} } \frac{\left (V^{\frac{7}{5}-1 } \right )^{V_{1} }_{0} }{\frac{-7}{5} +1}[/tex]

[tex]W=P_{0}V_{0}^{7/5} } \frac{\left (V^{\frac{-2}{5} } \right )^{V_{1} }_{0} }{\frac{-2}{5} }[/tex]

[tex]W=\frac{5}{2} P_{0}V_{0}^{7/5} }\left ( V_{0}^{-2/5} \right - V_{1}^{\frac{-2}{5} } } )[/tex]

Above is the work done by the gas on the container during the expansion

Part C-

It is known as a change in internal energy can be given as-

[tex]\bigtriangleup U=nC_{v} \bigtriangleup T[/tex]

Now find change in tempreture first,

[tex]\bigtriangleup T=T-T_{0}[/tex]

[tex]\bigtriangleup T= \frac{P_{1} V_{1} }{nR } - \frac{P_{0} V_{0} }{nR }[/tex]

[tex]\bigtriangleup T= \frac{1}{nR}\left ( P_{1} V_{1} \right-P_{0} V_{0} )[/tex]

[tex]\bigtriangleup T= \frac{1}{nR}\left ( P_{0} V_{0}^{7/5 }V_{1}^{-7/5} \right-P_{0} V_{0} )[/tex]

[tex]\bigtriangleup T= \frac{P_{0}}{nR}\left ( V_{0}^{7/5 }V_{1}^{-7/5} \right- V_{0} )[/tex]

Now Internal Energy,

[tex]\bigtriangleup U= n\frac{5R}{2} \frac{P_{0}}{nR}\left ( V_{0}^{7/5 }V_{1}^{-7/5} \right- V_{0} )[/tex]

[tex]\bigtriangleup U= \frac{5}{2} P_{0} \frac{P}\left ( V_{0}^{7/5 }V_{1}^{-7/5} \right- V_{0} )[/tex]

This is the final change in energy

For more detail follow the link:

https://brainly.com/question/14926413

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