Calculating the Value of Q for Constant Pressure Heat Transfer

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SUMMARY

The value of Q for the heating of 5 moles of an ideal gas from 300 K to 500 K at constant pressure is definitively 29.1 kJ, equivalent to 29100 J. In this scenario, the heat transferred (Q) is distinct from the change in internal energy (ΔU), as ΔU is influenced by both heat transfer and work done during expansion. The relationship between heat transfer and internal energy is clarified by the equation ΔU = Q - W, where W represents work done, which is zero in a constant volume situation. The discussion confirms that for constant pressure, Q equals the heat added, while ΔU varies based on the specific heat capacities C_p and C_v.

PREREQUISITES
  • Understanding of the ideal gas law
  • Familiarity with thermodynamic principles, specifically heat transfer
  • Knowledge of specific heat capacities (C_p and C_v)
  • Basic grasp of the first law of thermodynamics
NEXT STEPS
  • Study the derivation of the first law of thermodynamics
  • Learn about the differences between isochoric and isobaric processes
  • Explore the calculation of work done during gas expansion
  • Investigate the relationship between specific heat capacities and their applications in thermodynamics
USEFUL FOR

Students and professionals in thermodynamics, mechanical engineers, and anyone studying heat transfer processes in ideal gases will benefit from this discussion.

annalise17
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1. What is the value of Q in this problem? Is it a value that's stated in the question or does it need calculating?

A sample consisting of 5 moles of an ideal gas at a temperature of 300 K and a
pressure of 1.00 × 105 Pa is heated to a temperature of 500 K at constant pressure.
The amount of heat transferred to the gas is 29.1 kJ.

Homework Equations


No equations needed really, I just need to make sure I'm using the right values!


The Attempt at a Solution


Is Q = 29.1 kJ = 29100 J (or would that be U?)
Is Q the same value for constant volume and constant pressure while ΔU changes because ΔU = Q + W and there is no work in a constant volume situation?
 
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In this example the gas is at constant pressure which means that it will expand and therefore external work is done during the expansion.
All of the heat energy does not go to raise the internal energy (raise the temperature) of the gas
If the gas is heated at constant volume no external work is done ands all of the heat energy goes to internal energy and raise the temperature.
 
I got that bit, I just wasn't sure whether we were actually given Q or U but I've worked it out:

∆U = Q = 29100 J
∆U = C_v (T_2-T_1)
etc. Using Q as that gave me a sensible ratio of C_p to C_v so I'm happy!
 

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