Three moles of an ideal monatomic gas are at a temperature of 396 K. Then 2438 J of heat is added to the gas, and 897 J of work is done on it. What is the final temperature of the gas?
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ΔQ=ΔU+W
ΔU= ΔQ-W=2438-897 =1541 J
ΔU=νcΔT=ν(iR/2) ΔT
ΔT=2 ΔU/νiR =2•1541/3•3•8.31=41.2 K
ΔT= T₂-T₁
T₂=T₁+ΔT=396+41.2=437.2 K
To find the final temperature of the gas, we can use the first law of thermodynamics, which states that the change in internal energy (∆U) of a system is equal to the heat added (Q) minus the work done (W) on the system.
The equation for the first law of thermodynamics is:
∆U = Q - W
Since we are given the heat added (Q) and the work done (W), we can substitute the values into the equation and calculate the change in internal energy (∆U).
∆U = 2438 J - 897 J
∆U = 1541 J
Next, we need to relate the change in internal energy (∆U) to the change in temperature (∆T) of the gas. The relationship between ∆U and ∆T can be expressed using the equation:
∆U = n * Cv * ∆T
Where:
- ∆U is the change in internal energy.
- n is the number of moles of gas.
- Cv is the molar specific heat capacity at constant volume.
- ∆T is the change in temperature.
Since we are looking for the final temperature, we can rearrange the equation to solve for ∆T:
∆T = ∆U / (n * Cv)
The specific heat capacity at constant volume (Cv) for a monatomic gas can be calculated using the equation:
Cv = (3/2) * R
Where:
- R is the ideal gas constant (8.314 J/(mol·K)).
Now, we can calculate ∆T:
∆T = 1541 J / (3 mol * (3/2) * 8.314 J/(mol·K))
∆T ≈ 30.97 K
Therefore, the final temperature of the gas is approximately equal to 396 K + 30.97 K.
Final temperature = 426.97 K