Entropy: gas heated by resistor

AI Thread Summary
A current of 0.2 A flows through a 50 Ω resistor in an adiabatic vessel containing 3 moles of helium at an initial temperature of 27 °C. The electrical energy converted to heat raises the temperature of both the gas and the resistor, leading to calculations for internal energy and entropy changes. The initial calculations for temperature equilibrium were incorrect due to a miscalculation of the power dissipated by the resistor. After correcting the calculations, the final values for the internal energy change of helium and the entropy changes were approximately 982.1 J and 3.14 J/K, respectively. The discussion highlights the importance of accurate numerical calculations in thermodynamic problems.
JMatt7
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Homework Statement


A current ## I=0.2 A ## flows in a resistor ##R = 50 Ω## immersed in a rigid adiabatic vessel that contains ##n=3## moles of Helium. The initial temperature of the system is ##T_0 = 27 °C##. The resistor has a mass ## m = 10 g## and specific heat ## c = 0.2 (cal/K)/g ## and runs for ##t= 10 min##. Find:
  1. the internal energy change of He.
  2. the entropy change of He and the entropy change of the Universe.

Homework Equations


##P = i^2R##
##dS = \left( \frac {\delta Q} T \right)##

...

The Attempt at a Solution


I think I solved the problem, but I'm not completely sure of my reasoning, especially on the last question. Could you tell me what you think?

1) Helium is a noble gas, so I assumed it's an ideal gas. Therefore ## U = U(T) ##.
Since the volume is kept constant: ## ΔU = n c_V ΔT = n \frac 3 2 RΔT##. (Helium is monoatomic)
I calculated the temperature at equilibrium (##T_{eq}##) by knowing that the electrical energy is converted into heat that heats up both the resistor and the gas:
$$E_{in} = (mc + n c_v) (T_{eq} - T_0) $$
$$ I^2 R t + T_0 (mc + n c_v) = T_{eq} (mc + n c_v)$$
and hoping I converted all units accordingly:
$$T_{eq} = \frac {I^2 R t + T_0 (mc + n c_v)} {(mc + n c_v)} ≈ 431 K $$
Therefore: $$ΔU = n \frac 3 2 RΔT ≈ 4.89 kJ$$

2) Helium absorbs heat at constant volume, so ##TdS =\delta Q = nc_v dT##
So: $$ ΔS_{He} = \int_{T_0}^{T_{eq}} \frac {nc_v} T \, dT = n c_v \ln{\frac {T_{eq}} {T_0}} ≈13.5 J/K $$
As for the entropy change of the universe, I calculated the entropy change due to the heating of the resistor and added it to that of the gas:
$$ ΔS_R = \int_{T_0}^{T_{eq}} \frac {mc} T \, dT = mc \ln{\frac {T_{eq}} {T_0}} ≈3.02 J/K $$
$$ ΔS_{univ} = ΔS_{He} + ΔS_R ≈ 16.52 J/K$$

Since the energy flowing into the system comes only through electrical work, and the vessel is adiabatic, there are no other heat exchanges that could cause a change in entropy of the universe, right?
 
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Hello and welcome to PF!

Your method and resoning look very good to me. However, I'm getting different numbers.

For example, you have
##T_{eq} = \frac {I^2 R t + T_0 (mc + n c_v)} {(mc + n c_v)} ≈ 431 K ##
The formula looks correct, but I don't get 431 K. Can you tell us what numerical values you have for ##mc## for the resistor and ##nc_v## for the He?
 
TSny said:
The formula looks correct, but I don't get 431 K. Can you tell us what numerical values you have for ##mc## for the resistor and ##nc_v## for the He?
Sure. I might have been a bit sloppy with numbers. So using the given data:
##mc = 10 g ~ 0.2~ \frac {cal} {K~g} = 2~ \frac {cal} K ##
##nc_v = 3~ mol~ \frac {3}{2}~ R = 3~mol~ \frac {3}{2}~ 1.987~ \frac {cal}{K ~mol} ≈ 8.94 \frac {cal} K##
Therefore, since ## 1 ~cal = 4.18 ~J##:
$$T_{eq} = \frac {6000 ~J + 300.15~ K (2+8.94)~ 4.18 \frac J K} {(2+8.94)~ 4.18 \frac J K} ≈ \frac {6000 ~J + 13727.5 ~J } {45.74 \frac J K} ≈ 431.3 K$$
 
OK. The problem appears to be with your calculation of ##I^2Rt##. Did you square the current?
 
TSny said:
OK. The problem appears to be with your calculation of ##I^2Rt##. Did you square the current?
Ahaha. Yes, you're right. Apparently I didn't. So ##I^2Rt = 1200 ~J## and therefore:
##T_{eq} ≈ 326.4~ K##
##ΔU ≈ 982.1 ~J##
##ΔS_{He} ≈ 3.14 ~J/K##
##ΔS_R ≈ 0.7 ~J/K##
 
OK. I get about 987 J for ΔUHe, but that's close to what you get.
 
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