Calculating Internal Energy Change in Isobaric Expansion

Click For Summary
In an isobaric expansion of an ideal gas at 230 K and 3.65 kPa, the volume increases from 1.7 m³ to 6.8 m³, with 25.1 kJ of heat transferred to the gas. To calculate the change in internal energy, the formula used is ΔU = q + w, where q is the heat added and w is the work done by the gas. The work done can be calculated using the formula w = -P * ΔV, resulting in a specific value based on the pressure and change in volume. The heat added (25.1 kJ) contributes positively to the internal energy, while the work done by the gas during expansion is subtracted. Ultimately, the change in internal energy reflects the balance between the heat added and the work performed by the gas.
nemzy
Messages
124
Reaction score
0
question: An ideal gas initially at 230 K undergoes an isobaric expansion at 3.65 kPa.

If the volume increases from 1.7 m^3 to 6.8 m^3 and 25.1 kJ is transferred to the gas by heat, what is the change in its internal energy?


i know that isobaric expansion means that the pressure is constant. however, i don't know how to calculate the internal energy. my proff went over this part very briefly and the book doesn't cover much about this topic.
 
Physics news on Phys.org
Use the specific heat for an ideal gas at constant pressure.
 
so internal energy = q+w

work = -p*change in volume

which is (3.65kpa*1000)(6.8m^3-1.70m^3)

and q= n*Cp*change in T

and Cp = 20.775 J/mol*K

i know the initial T, and could find the final T
n is also easy to find

however, what does the 25.1 kj transferred to the gas by heat mean and how does it relate to this problem?
 
Last edited:
Heat (thermal energy) is being added to the system and the gas expands. As the gas expands it does work. The change in internal energy will be the difference between the heat added and the work done by the gas.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 81 ·
3
Replies
81
Views
6K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
5
Views
2K