Question

. A. find the change in temperature ∆� of a monatomic ideal gas if it absorbs
9750J at a constant volume of 2.40L
b. what is the change in pressure, ∆�?
c. if the system is an ideal diatomic gas, find the change in its temperature.

Answer 1

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.