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An ideal gas with γ = 1.67 has an initial temperature of 0°C, initial volume of 10.0 liters, and initial pressure of 1.00 atm. Then the gas is expanded adiabatically to a volume of 10.4 liters. What is the new temperature? (1 point)

Respuesta :

Answer: T = 9.74°C

Explanation: An ideal gas in a quasi-static adiabatic process follows the equation: [tex]pV^{\gamma} = constant[/tex].

So:

[tex]1.10^{1.67} = constant[/tex]

constant = 46.7735

Adiabatic conditions can be written as:

[tex]TV^{\gamma-1}=constant[/tex]

Then, new temperature is

[tex]T.(10.4)^{1.67-1}=46.7735[/tex]

[tex]T.(10.4)^{0.67}=46.7735[/tex]

[tex]T = \frac{46.7735}{4.802}[/tex]

T = 9.74°C

The new temperature is 9.74°C.

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