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3 months ago

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The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

n. 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.

1 answer:

Yuliya22 [3.3K]3 months ago

6 0

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

It is known that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v are constant on both sides. Therefore we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let the final pressure be P and the temperature $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide equation (2) by equation (3)

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

$T_1 {f} = 1.48 T_2 {f}$

thus, the left side temperature equals 1.48 times the right side temperature

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