Let's tackle each part of the problem step-by-step:
Part A: Change in Temperature for Monatomic Ideal Gas
For a monatomic ideal gas, the molar heat capacity at constant volume (\(C_V\)) is given by:
\[ C_V = \frac{3}{2}R \]
where \(R\) is the ideal gas constant, approximately \(R = 8.314 , \text{J/(mol·K)}\).
To find the change in temperature (\(\Delta T\)) given the heat absorbed (\(Q\)) at constant volume, we can use the formula:
\[ Q = n C_V \Delta T \]
Rearranging gives:
\[ \Delta T = \frac{Q}{n C_V} \]
To find \(n\) (the number of moles), we can use the ideal gas law:
\[ PV = nRT \]
Since the pressure \(P\) is not provided, let's assume it is at standard conditions for simplicity, which is \(P = 101.3 , \text{kPa}\) or \(101300 , \text{Pa}\). We will have to convert the volume from liters to cubic meters:
\[ V = 2.40 , \text{L} = 2.40 \times 10^{-3} , \text{m}^3 \]
Using the ideal gas law, we can find \(n\):
\[ n = \frac{PV}{RT} = \frac{(101300 , \text{Pa})(2.40 \times 10^{-3} , \text{m}^3)}{(8.314 , \text{J/(mol·K)})(298 , \text{K})} \]
Calculating this gives:
\[ n \approx \frac{243.12}{2477.572} \approx 0.098 , \text{moles} \]
Now we can compute \(\Delta T\):
\[ C_V = \frac{3}{2}(8.314) \approx 12.471 , \text{J/(mol·K)} \]
Now substituting \(Q\), \(n\), and \(C_V\):
\[ \Delta T = \frac{9750 , \text{J}}{0.098 , \text{moles} \times 12.471 , \text{J/(mol·K)}} \approx \frac{9750}{1.222} \approx 7988.36 , \text{K} \]
Part B: Change in Pressure (\(∆P\))
Using the ideal gas law, we can find the initial pressure \(P_i\) and the final pressure \(P_f\), where temperature has changed but volume remains constant.
Let \(P_i = \frac{nRT_i}{V}\) and \(P_f = \frac{nRT_f}{V}\).
The change in pressure is:
\[ \Delta P = P_f - P_i = \frac{nR(T_f - T_i)}{V} \]
Substituting \(T_f = T_i + \Delta T\) gives:
\[ \Delta P = \frac{nR \Delta T}{V} \]
Now substituting the already calculated values into this formula:
\[ \Delta P = \frac{(0.098)(8.314)(7988.36)}{2.40 \times 10^{-3}} \approx \frac{0.98 \times 8.314 \times 7988.36}{2.40 \times 10^{-3}} \approx \frac{7996.56}{0.0024} \approx 3335473.81 , \text{Pa} \approx 3.335 , \text{MPa} \]
Part C: Change in Temperature for Ideal Diatomic Gas
For a diatomic ideal gas, the molar heat capacity at constant volume (\(C_V\)) is:
\[ C_V = \frac{5}{2}R \]
Using the same formula for \(\Delta T\):
\[ \Delta T = \frac{Q}{n C_V} \]
For a diatomic gas:
\[ C_V = \frac{5}{2}(8.314) \approx 20.785 , \text{J/(mol·K)} \]
Calculating \(\Delta T\):
\[ \Delta T = \frac{9750 , \text{J}}{0.098 , \text{moles} \times 20.785 , \text{J/(mol·K)}} \]
Calculating:
\[ \Delta T = \frac{9750}{2.03123} \approx 4800.37 K \]
Summary of Results
- Part A: \(\Delta T \approx 7988.36 , \text{K}\) for monatomic gas.
- Part B: \(\Delta P \approx 3.335 , \text{MPa}\).
- Part C: \(\Delta T \approx 4800.37 , \text{K}\) for diatomic gas.