The work done by one mole of a monatomic ideal gas () in expanding adiabatically is 825 J. The initial temperature and volume of the gas are 393 K and 0.100 m3. Obtain (a) the final temperature and (b) the final volume of the gas.
To solve this problem, we will use the first law of thermodynamics for an adiabatic process:
\(W = \Delta U + \Delta E_k + \Delta E_p\)
Where:
- \(W\) is the work done on or by the gas
- \(\Delta U\) is the change in internal energy of the gas
- \(\Delta E_k\) is the change in kinetic energy of the gas
- \(\Delta E_p\) is the change in potential energy of the gas
Since the gas is monatomic, it does not have any potential energy (\(\Delta E_p = 0\)) and the change in kinetic energy (\(\Delta E_k = 0\)).
So, the equation simplifies to:
\(W = \Delta U\)
Now, the work done by the gas is given as 825 J. We know that for an ideal gas, the internal energy (\(\Delta U\)) is related to the temperature change (\(\Delta T\)) using the equation:
\(\Delta U = nC_v \Delta T\)
Where:
- \(n\) is the number of moles of the gas (given as 1 mole)
- \(C_v\) is the molar heat capacity at constant volume for the gas
For a monatomic ideal gas, \(C_v\) is given as \(3/2 R\), where \(R\) is the ideal gas constant.
Let's substitute these values into the equation:
825 J = (1 mol)(3/2 R)(\(\Delta T\))
We can rearrange the equation to solve for \(\Delta T\):
\(\Delta T = \frac{825 J}{(1 \text{ mol})(3/2 R)}\)
Next, we can calculate the value of \(\Delta T\) by substituting the value of the ideal gas constant, \(R\), which is approximately 8.314 J/(mol·K):
\(\Delta T = \frac{825 J}{(1 \text{ mol})(3/2)(8.314 J/(mol·K))}\)
After evaluating this expression, we will get the value of \(\Delta T\).
(a) To find the final temperature, we need to add \(\Delta T\) to the initial temperature:
\(T_{\text{final}} = T_{\text{initial}} + \Delta T\)
Substituting in the given values for the initial temperature and the calculated value of \(\Delta T\), we can find the final temperature.
(b) To find the final volume, we can use the equation for an adiabatic process in an ideal gas:
\(T_1V_1^{\gamma - 1} = T_2V_2^{\gamma - 1}\)
Where:
- \(T_1\) and \(V_1\) are the initial temperature and volume
- \(T_2\) and \(V_2\) are the final temperature and volume
- \(\gamma\) is the heat capacity ratio, which is \(C_p/C_v\)
For a monatomic ideal gas, \(\gamma\) is equal to 5/3.
By rearranging this equation, we can solve for the final volume, \(V_2\):
\(V_2 = V_1\left(\frac{T_1}{T_2}\right)^{(1-\gamma)/\gamma}\)
Substituting in the given values for the initial temperature and volume, the calculated value of the final temperature, and the value of \(\gamma\), we can find the final volume.