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The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partition. Both sides contain one mole of a monatomic ideal gas (γ=5/3) with the initial temperature being 525 K on the left and 275 K on the right. The partition is then allowed to move slowly to the right until the pressures on each side of the partition are the same. Find the final temperatures on the left and right side.

Respuesta :

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

[tex]\gamma = 5/3[/tex]

T1 = 525 K

T2 = 275 K

We know that

[tex]P_1 = \frac{nRT_1}{v}[/tex]

[tex]P_2 = \frac{nrT_2}{v}[/tex]

n and v remain same at both side. so we have

[tex]\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}[/tex]

[tex]P_1 = \frac{21}{11} P_2[/tex] ..............1

let final pressure is P and temp  [tex]T_1 {f} and T_2 {f}[/tex]

[tex]P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}[/tex]

[tex]P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}[/tex] ..................2

similarly

[tex]P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}[/tex] .............3

divide 2 equation by 3rd equation

[tex]\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}[/tex]

[tex]T_1 {f} = 1.48 T_2 {f}[/tex]

thus, temperature on left side is 1.48 times the temperature on right

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