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2 years ago

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6.1-2. Diffusion of CO, in a Binary Gas Mixture. The gas CO2 is diffusing at stcady state through a tube 0.20 m long having a di

ameter of 0.01 m and containing N2 at 298 K. The total pressure is constant at 101.32 kPa. The partial pressure of CO2 at one end is 456 mm Hg and 76 mm Hg at the other end. The diffusivity Dan is 1.67 x 10-5 m2/s at 298 K. Calculate the flux of CO2 in cgs and SI units for equimolar counterdiffusion

1 answer:

zzz [600]2 years ago

4 0

Answer:

Heat flux of CO₂ in cgs

= 170.86 x 10⁻⁹ mol / cm²s

SI units

170.86 x 10⁻⁸ kmol/m²s

Explanation:

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A rich industrialist was found murdered in his house. The police arrived at the scene at 11:00 PM. The temperature of the corpse

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The dude was killed around 6:30PM

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Newton's law of cooling states:

$T = T_m + (T_0-T_m)e^{kt}$

where,

$T_0$ = initial temp

$T_m$ = temp of room

$T$ = temp after t hours

$k$ = how fast the temp is changing

$t$ = time (hours)

$T_0 = 31$ because the body was initlally 31ºC when the police found it

$T_m = 22$ because that was the room temp

$T = 30$ because the body temp drop to 30ºC after 1 hour

t = 1 because that's the time it took for the body temp to drop to 30ºC

$k=???$ we don't know k so we must solve for this

rearrange the equation to solve for k

$T = T_m + (T_0-T_m)e^{kt}$

$T - T_m= (T_0-T_m)e^{kt}$

$\frac{T - T_m}{(T_0-T_m)}= e^{kt}$

$ln(\frac{T - T_m}{T_0-T_m})=kt$

$\frac{ln(\frac{T - T_m}{T_0-T_m})}{t}=k$

plug in the numbers to solve for k

$k = \frac{ln(\frac{T - T_m}{T_0-T_m})}{t}$

$k = \frac{ln(\frac{30 - 22}{31-22})}{1}$

$k=ln(\frac{8}{9})$

Now that we know the value for k, we can find the moment the murder occur. A crucial information that the question left out is the temperature of a human body when they're still alive. A living human body is about 37ºC. We can use that as out initial temperature to solve this problem because we can assume that the freshly killed body will be around 37ºC.

$T_0 = 37$ because the body was 37ºC right after being killed

$T_m = 22$ because that was the room temp

$T = 31$ because the body temp when the police found it

$k=ln(\frac{8}{9})$ we solved this earlier

t = ??? we don't know how long it took from the time of the murder to when the police found the body

Rearrange the equation to solve for t

$T = T_m + (T_0-T_m)e^{kt}$

$T - T_m= (T_0-T_m)e^{kt}$

$\frac{T - T_m}{(T_0-T_m)}= e^{kt}$

$ln(\frac{T - T_m}{T_0-T_m})=kt$

$\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}=t$

plug in the values

$t=\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}$

$t=\frac{ln(\frac{31 - 22}{37-22})}{ln(8/9)}$

$t=\frac{ln(3/5)}{ln(8/9)}$

$t=\frac{ln(3/5)}{ln(8/9)}$

t ≈ 4.337 hours from the time the body was killed to when the police found it.

The police found the body at 11:00PM so subtract 4.337 from that.

11 - 4.33 = 6.66 ≈ 6:30PM

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