Entropy change for isobaric heating

AI Thread Summary
The discussion focuses on calculating the entropy change (ΔSsurr) for a mole of nitrogen heated isobarically from 25°C to 100°C at a constant pressure of 1 atm. The user calculates ΔS using the formula ΔS = Q/T, resulting in -5.85 J/K, which contrasts with the lecture notes' value of -22 J/K. The user questions the discrepancy, suggesting a possible error in the lecture notes, specifically regarding the temperature used in the calculation. The conversation highlights the importance of using the correct temperature in entropy calculations and clarifies the confusion over the differing results. Accurate calculations are essential for understanding thermodynamic principles.
Alvine
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Hi guys, is this right?

Homework Statement


A mole of nitrogen is heated at constant external pressure of 1 atm from 25° C to 100° C. The heat capacity of nitrogen is 29.1 J/mol·K.

Calculate ΔSsurr




2. The attempt at a solution
delta_S=Q/T = -1*29.1*75/373=-5.85 J/K

(heat transferred to the system divided by T_surr which we assume to be equal to T_final)

The solution in my lecture notes says -22 J/K, I can't see where that came from.
 
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even if they heat it "infinitely slow", it only goes up to about 6½ J/K ,
no-where near 22 J/K .
{ they must've divided by 100 degrees C by mistake , rather than 373 K }
 
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