Molar Specific Heat of a Gas in Terms of R & s

AI Thread Summary
The discussion focuses on determining the molar specific heat, C_v, of a gas using the equipartition theorem, which states that each degree of freedom contributes an average energy of 1/2 k_B*T. Participants clarify that the molar specific heat can be expressed in terms of the number of degrees of freedom, s, and the gas constant, R. For example, a monatomic molecule has three translational degrees of freedom, leading to a specific heat of C_v = (3/2)k_BT. Additionally, the discussion includes a practical example involving cis-2-butene, where the molar specific heat is given, prompting a calculation of its energetically accessible degrees of freedom. The conversation emphasizes the relationship between degrees of freedom and molar specific heat in gases.
doggieslover
Messages
34
Reaction score
0
Part A
Using the equipartition theorem, determine the molar specific heat, C_v, of a gas in which each molecule has s degrees of freedom.
Express your answer in terms of R and s.

Okay, I know that the equipartition theorem is 1/2k_B*T

and molar specific heat is C_v= (1/n)(dU/dT)

But I don't know where to go from here, please help?

Part B

Given the molar specific heat C_v of a gas at constant volume, you can determine the number of degrees of freedom s that are energetically accessible.

For example, at room temperature cis-2-butene, \rm C_4 H_8, has molar specific heat C_v=70.6\;{\rm \frac{J}{mol \cdot K}}. How many degrees of freedom of cis-2-butene are energetically accessible?
Express your answer numerically to the nearest integer.
 
Physics news on Phys.org
doggieslover said:
But I don't know where to go from here, please help?
A monatomic molecule has three (translational) degrees of freedom and molar specific heat ##C_V=\frac{3}{2}k_BT.## Does this help?
 
doggieslover said:
Okay, I know that the equipartition theorem is 1/2k_B*T
Your confusion arises because ##\frac 12 k_{\rm B}T## isn't the equipartition theorem. Theorems are generally statements whereas ##\frac 12 k_{\rm B}T## by itself is akin to a single word.

The principle of equipartition of energy states that each degree of freedom has, on average, an associated energy per molecule of ##\frac 12 k_{\rm B}T##.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top