Chem Question

[swifteagle43]

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The average molecular kinetic energy of a sample of carbon dioxide is found to be the same as that of a sample of neon. How do the temperatures of the samples compare?

The answer is: The temperatures of the two gases are the same.

I was trying to figure out how this could be that something with greater molecular volume can be at the same speed but not have more temperature(energy). I know the equation is 3/2RT = K.E is proportional to V^2. How can mass not be introduced into this.... makes no sense

especially since the effusion law states that lighter gasses leak out faster because they move around faster....

 

[estairella]

KE = 1/2mv^2

That's where your mass is taken into account.

 

[CytochromeP450]

KE = 1/2mv^2

That's where your mass is taken into account.

I have sturggled with this one and it took me awhile to get it during ugrad. Basically, you see that particle with different masses can have the same K.E.. In that respectively, K.E. of the larger particle in compenstate by greater mass while the K.E. of the smaller particle is compenstate by its velocity.

 

[wisguy]

0.5*m*v(avg)^2 = 1.5*k*T = KE(avg) at a given temp.

for gases a and b,

KE(a) = KE(b) = 1.5*k*T = constant

thus, 0.5*m(a)*v(a)^2 = 0.5*m(b)*v(b)^2

in your question, your assumption that molecules with the same average KE have the same average velocity is not valid. The KE is the same for two gases at any given T, but if the masses differ then the avg speed must also be different. In this way, the law of effusion is maintained; light gas molecules move faster than heavy ones.

 

[swifteagle43]

I have sturggled with this one and it took me awhile to get it during ugrad. Basically, you see that particle with different masses can have the same K.E.. In that respectively, K.E. of the larger particle in compenstate by greater mass while the K.E. of the smaller particle is compenstate by its velocity.

wow, that is clear

 

[swifteagle43]

0.5*m*v(avg)^2 = 1.5*k*T = KE(avg) at a given temp.

for gases a and b,

KE(a) = KE(b) = 1.5*k*T = constant

thus, 0.5*m(a)*v(a)^2 = 0.5*m(b)*v(b)^2

in your question, your assumption that molecules with the same average KE have the same average velocity is not valid. The KE is the same for two gases at any given T, but if the masses differ then the avg speed must also be different. In this way, the law of effusion is maintained; light gas molecules move faster than heavy ones.

nice

 

[booji]

This concept actually is one of the clauses of the KMT (Kinetic Molecular Theory) and I am pretty sure that it holds true only for ideal gases. The real gases will show Van Der Waals' deviation.