A monatomic ideal gas that is initially at a pressure of 1.50 * 105 Pa and has a volume of 0.08 m3 is compressed adiabatically to a volume of 0.0400 m3 . (a) What is the final pressure? (b) How much work is done by the gas? (c) What is the ratio of the final temperature of the gas to its initial temperature? Is the gas heated or cooled by this compression

Answer :

Answer:

1.571 is the required ratio

Solution:

As per the question:

Pressure, P = [tex]1.50\times 10^{5}\ Pa[/tex]

Volume, V = [tex]0.08\ m^{3}[/tex]

Volume,V' = [tex]0.04\ m^{3}[/tex]

(a) To calculate the final pressure:

In adiabatic compression:

[tex]PV^{\gamma} = P'V'^{\gamma}[/tex]

[tex]\gamma = 1.67[/tex]

[tex]P' = (\frac{0.04}{0.08})^{1.67}\times 1.50\times 10^{5}[/tex]

[tex]P' = 4.713\times 10^{5}\ Pa[/tex]

(b) From the eqn of ideal gas:

PV = nRT

We can write:

[tex]\frac{P'V'}{PV} = \frac{T'}{T}[/tex]

where

T' = Final temperature

T = Initial temperature

Thus

[tex]\frac{4.713\times 10^{5}\times 0.04}{1.50\times 10^{5}\times 0.08} = \frac{T'}{T}[/tex]

[tex]\frac{T'}{T} = 1.571[/tex]

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