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Basile [38]
2 years ago
8

Consider two cylinders of gas identical in all respects except that one contains oxygen o2 and the other helium he. both cylinde

rs initially contain the same volume of gas at 0°c and 1 atm of pressure and are closed by a movable piston at one end. both gases are now compressed adiabatically to one-third their original volume. (a) which gas will show the greater temperature increase?
Physics
1 answer:
kipiarov [429]2 years ago
5 0
1) For a reversible adiabatic process, the ideal gas law can be written also as 
TV^{\gamma-1}=cost.
or equivalently
T_iV_i^{\gamma-1}=T_fV_f^{\gamma-1}
where T_i and T_f are the initial and final temperature of the gas, and V_i and V_f are its initial and final volume. \gamma is the adiabatic index, given by
\gamma= \frac{C_P}{C_V}= \frac{f+2}{f}
where f is the number of degrees of freedom of the molecule. For helium, which is monoatomic gas, we have f=3, therefore
\gamma_{He}= \frac{5}{3}
Instead, for oxigen (O_2) which is a diatomic gas, we have f=5, therefore
\gamma_{O_2}= \frac{7}{5}

2) Using the initial relationship written at point 1), we can now calculate the increase in temperature for both gases. First of all, let's rewrite the initial equation as:
T_f=( \frac{V_i}{V_f})^{\gamma-1} T_i
And  since we know that both gases are compressed to one-third of their original volume, i.e.
V_f= \frac{1}{3}V_i
this means
T_f=3^{\gamma-1} T_i

So now we can calculate the final temperature for each gas, since we know the initial temperature: T_i=0^{\circ}C=273~K 

- For helium: 
T_f=3^{ \frac{5}{3}-1} T_i = 3^{ \frac{2}{3} } \cdot 273~K = 568~K

- For oxigen:
T_f=3^{ \frac{7}{5}-1 } T_i = 3^{ \frac{2}{5} }\cdot 273~K = 424~K

So, helium shows the greater temperature increase.
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The question is not clear and the complete clear question is;

The velocity profile in fully developed laminar flow In a circular pipe of inner radius R = 2 cm, in m/s, is given By u(r) = 4(1 - r²/R²). Determine the average and maximum Velocities in the pipe and the volume flow rate.

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Explanation:

A) We are given that;

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4(0 - (2r/R²)) = 0

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V_max = 4(1 - 0²/R²) = 4 - 0 = 4 m/s

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