Recent content by VSayantan

Yes. I tried to convince myself it was the closest answer (which would be a 100 times different than ##13.82~k_B##) before you suggested the one state population scenario.

Yes. You're absolutely right @Chestermiller :smile:

And you argued that that it is most probable for the bosons to occupy the same state. Which gives only one most probable arrangement, when there is no constraint of temperature involved. So, I think you're right and zero is the correct choice, I think.

Isn't your unit for ##T## - ##K##?

There are six options given ##13.82~k_B## ##693.1~k_B## ##zero## ##1000~k_B## ##5909.693~k_B## ##6909~k_B## And no the question does not ask for entropy as a function of temperature. Thanks for your suggestion @PeterDonis :bow:

For the bosons I've already calculated the entropy.

The question clearly says "the temperature of ##n=19~\rm {mol}## of argon gas ..." why did you use ##n=1##? How did you find the change in temperature to add with?

How did you obtain the value ##32.14##? So your initial equation was wrong! It should be $$Q=m\cdot {C_p} \cdot {(T_f \sim T_i)}$$

So, ##\Delta T## is change in temperature, right? Now, say, initially the temperature is ##T_i## and finally ##T_f##. What is the change in temperature?

cool. Write that in symbols.

Yep. Me too. But what is it? Can you elaborate it?

What is ##\Delta T##?

Is it possible that since a given energy state can accommodate two spin-half particles, at most, there is only one to distribute these ##2N## particles in ##N## states?

Homework Statement A system having ##N## non-degenerate energy eigenstates populated by##N## identical spin-zero particles and ##2N## identical spin-half particles. There are no interaction between any of these particles. If ##N=1000## what is the entropy of the system? Homework Equations...

OK, I think I've found a way. The energy required to increase the temperature of the gas from ##T_0~K## to ##{(T_0+1)}~K## is $$\Delta {U_{gas}}={U_{gas}}_i - {U_{gas}}_f$$ Which simplifies to $$\Delta {U_{gas}}={\frac 3 2}RT_0-{\frac 3 2}R{(T_0+1)}$$ That is $$\Delta {U_{gas}}=-{\frac 3 2}R$$...