Entropy change for isobaric heating

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SUMMARY

The discussion revolves around calculating the change in entropy for a mole of nitrogen heated isobarically from 25°C to 100°C at a constant external pressure of 1 atm. The heat capacity of nitrogen is specified as 29.1 J/mol·K. The user calculated ΔSsurr as -5.85 J/K, but noted a discrepancy with the lecture notes, which state the value as -22 J/K. The user suspects an error in the lecture notes, specifically regarding the temperature conversion from Celsius to Kelvin.

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  • Understanding of the first law of thermodynamics
  • Knowledge of heat capacity and its application in thermodynamic calculations
  • Familiarity with the concept of entropy and its significance in thermodynamics
  • Ability to convert temperatures between Celsius and Kelvin
NEXT STEPS
  • Review the principles of isobaric processes in thermodynamics
  • Study the calculation of entropy changes in thermodynamic systems
  • Learn about the relationship between heat transfer and temperature in entropy calculations
  • Examine common mistakes in temperature conversions in thermodynamic equations
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Students studying thermodynamics, educators teaching heat transfer concepts, and anyone involved in physical chemistry or engineering disciplines focusing on thermodynamic calculations.

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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