8.31 Wh (watthours) heat is transmitted to 1 mole single atomic ideal gas. The gas temperature is 27°C and the pressure is constant during the process (slow process). What is the temperature? What is the change of the internal energy of the gas? What is the work done during the gas is surrounding?
http://uni-obuda.hu/users/racz.ervin/Problems_thermodynamics.pdf
What about the answer?
I didn't understand any thing
The answer is given in the link.
To find the temperature, change of internal energy, and work done, we can use the first law of thermodynamics, also known as the law of energy conservation:
ΔU = Q - W
Where:
ΔU is the change in internal energy of the gas
Q is the heat transferred to the gas
W is the work done by the gas
Given:
Q = 8.31 Wh (heat)
We need to convert this to joules since the unit of energy in thermodynamics is typically joules.
1 Wh = 3600 J
So, Q = 8.31 Wh * 3600 J/Wh = 29,916 J
Since the ideal gas is single atomic, it has only 3 degrees of freedom. The internal energy of a monoatomic ideal gas is given by:
U = (3/2) nRT
Where:
n is the number of moles
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
Since the process is slow, the pressure remains constant. Therefore, the temperature will also remain constant.
To find the temperature, we need to convert the given temperature from Celsius to Kelvin by adding 273.15:
T = 27°C + 273.15 = 300.15 K
Substituting these values into the equation for the internal energy:
U = (3/2) * 1 * R * 300.15 K = 450.225 J/mol
Now, we can calculate the change in internal energy (ΔU):
ΔU = Q - W
Since the process is slow and the temperature is constant, the internal energy does not change, so ΔU = 0. Therefore, we can solve for W:
W = Q = 29,916 J
Lastly, we can substitute the values into the equation to find the work done (W):
W = Q = 29,916 J
To summarize:
- The temperature remains constant at 300.15 K
- The change in internal energy (ΔU) is 0
- The work done (W) is 29,916 J