One mole of ideal diatomic gas expands... final temperature will be (close to) ___ $^\circ$C.
Show Hint
- -56
- -30
- -189
- -117
The Correct Option is A
Solution and Explanation
In this problem, we are dealing with one mole of an ideal diatomic gas. We need to determine the final temperature after expansion, given that the process is adiabatic. This means no heat is transferred into or out of the system. Let's solve this step-by-step using thermodynamics principles.
For an adiabatic process, the relation between the initial and final states can be given by:
\(PV^\gamma = \text{constant}\)
where:
- \(P\) is the pressure
- \(V\) is the volume
- \(\gamma\) is the heat capacity ratio (specific heat ratio), for diatomic gases \(\gamma \approx \frac{7}{5} = 1.4\)
The transformation rule for temperature during an adiabatic process for an ideal gas is:
\(\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}\)
where:
- \(T_1\) and \(T_2\) are the initial and final temperatures, respectively
- \(V_1\) and \(V_2\) are the initial and final volumes, respectively
Given that the expansion is a doubling of the volume at a constant pressure, we have:
- \(V_2 = 2V_1\)
Substituting this into the equations, we have:
\(\frac{T_2}{T_1} = \left(\frac{V_1}{2V_1}\right)^{\gamma-1} = \left(\frac{1}{2}\right)^{0.4}\)
Calculate the above expression:
\(\left(\frac{1}{2}\right)^{0.4} \approx 0.742\)
Thus, we have:
\(T_2 = T_1 \times 0.742\)
Let's assume the initial temperature \(T_1\) to be in Kelvin;
Converting from Celsius (assuming a standard initial temperature, like 27°C which is approximately 300 K for an initial context):
\(300 \times 0.742 = 222.6 \, \text{K}\)
Convert the final temperature to Celsius:
\(222.6 \, \text{K} - 273 \approx -50.4 \,^\circ\text{C}\)
The closest option available is \(-56 \,^\circ\text{C}\), which is the correct answer.
Therefore, upon the adiabatic expansion, the final temperature of the gas is approximately -56°C.
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