Consider two different containers, each filled with of . One of the containers is rigid and has constant volume. The other container is flexible (like a balloon) and is capable of changing its volume to keep the external pressure and internal pressure equal to each other. If you raise the temperature in both containers, what happens to the pressure and density of the gas inside each container? Assume a constant external pressure.

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lucianoangelini92 lucianoangelini92 · Answer 1 · 2019-11-08T15:40:22+01:00

Answer:

In the rigid container : pressure increases when temperature increases and density remains constant

In the flexible container: pressure remains constant and density decreases when temperature increases

Explanation:

Using the Ideal gas law

[tex]P*V=n*R*T[/tex]

P= absolute pressure, V=volume , n= number of moles (mass) , R=constant , T= absolute temperature

And

[tex]D= m/V[/tex]

D= Density , m = mass , V= volume

In both containers, since they are sealed, the mass is kept inside and remains constant --> n= constant and m= constant

- In the rigid container , V = constant , therefore

[tex]P*V=n*R*T\\\\P=n*R*T/V = (n*R/V) * T = constant * T\\\\P = constant * T[/tex]

And thus absolute pressure increases with absolute temperature

Regarding density

[tex]D= m/V = constant 1 /constant 2= constant 3[/tex]

Thus density remains constant

- In the flexible container , P = constant , therefore

[tex]P*V=n*R*T\\\\V=n*R*T/P = (n*R/P) * T = constant * T\\\\V = constant * T[/tex]

but also

[tex]D= m/V = constant 1 /(constant 2* T)= constant 3 / T\\\\D = constant 3* T[/tex]

And thus density decreases with temperature