If two separate systems, one containing a monatomic gas, and the other a diatomic gas undergo an adiabatic process from V1 to V2, which would have the lower final pressure, p2? What would be the ratio of the two final pressures? Give your answer in terms of V1 and V2. Assume the gases are at approximately room temperature.

To compare the final pressures of a monatomic gas and a diatomic gas undergoing an adiabatic process, we can use the following relationship for adiabatic processes:

\[
P V^\gamma = \text{constant}
\]

where \( \gamma = \frac{C_p}{C_v} \). For an ideal monatomic gas, \( \gamma = \frac{5}{3} \), and for an ideal diatomic gas, \( \gamma = \frac{7}{5} \).

### Final Pressures

For each gas, we can express the initial and final states in terms of pressure, volume, and \(\gamma\):

1. **Monatomic Gas:**
   \[
   P_1 V_1^{\gamma} = P_2 V_2^{\gamma}
   \]
   Rearranging gives:
   \[
   P_2^{\text{mono}} = P_1 \left(\frac{V_1}{V_2}\right)^{\gamma_{\text{mono}}}
   \]
   \[
   P_2^{\text{mono}} = P_1 \left(\frac{V_1}{V_2}\right)^{\frac{5}{3}}
   \]

2. **Diatomic Gas:**
   \[
   P_1 V_1^{\gamma} = P_2 V_2^{\gamma}
   \]
   Rearranging gives:
   \[
   P_2^{\text{di}} = P_1 \left(\frac{V_1}{V_2}\right)^{\gamma_{\text{di}}}
   \]
   \[
   P_2^{\text{di}} = P_1 \left(\frac{V_1}{V_2}\right)^{\frac{7}{5}}
   \]

### Comparison of Final Pressures

To find which gas has a lower final pressure, we need to analyze the exponents:

- For the monatomic gas: \( \left(\frac{V_1}{V_2}\right)^{\frac{5}{3}} \)
- For the diatomic gas: \( \left(\frac{V_1}{V_2}\right)^{\frac{7}{5}} \)

Since \( \frac{7}{5} > \frac{5}{3} \) (because \( 7/5 \approx 1.4 \) and \( 5/3 \approx 1.67 \)), the diatomic gas's final pressure will be lower if the same initial pressure \( P_1 \) is applied to both. Thus:

\[
P_2^{\text{di}} < P_2^{\text{mono}}
\]

### Ratio of Final Pressures

The ratio of the final pressures can be expressed as:

\[
\frac{P_2^{\text{mono}}}{P_2^{\text{di}}} = \frac{P_1 \left(\frac{V_1}{V_2}\right)^{\frac{5}{3}}}{P_1 \left(\frac{V_1}{V_2}\right)^{\frac{7}{5}}} = \left(\frac{V_1}{V_2}\right)^{\frac{5}{3} - \frac{7}{5}}
\]

Calculating the exponent:

\[
\frac{5}{3} - \frac{7}{5} = \frac{25}{15} - \frac{21}{15} = \frac{4}{15}
\]

Thus, the ratio of the final pressures is:

\[
\frac{P_2^{\text{mono}}}{P_2^{\text{di}}} = \left(\frac{V_1}{V_2}\right)^{\frac{4}{15}}
\]

### Conclusion

1. The diatomic gas has the lower final pressure \( P_2^{\text{di}} < P_2^{\text{mono}} \).
2. The ratio of the final pressures is \( \frac{P_2^{\text{mono}}}{P_2^{\text{di}}} = \left(\frac{V_1}{V_2}\right)^{\frac{4}{15}} \).

Would you like to explore this further or have any questions? Here are some related topics to consider:

1. How does temperature affect the pressure in an adiabatic process?
2. What is the significance of the adiabatic process in thermodynamics?
3. How do real gases differ from ideal gases in adiabatic processes?
4. What are the implications of this analysis in practical applications like engines?
5. How does the value of \(\gamma\) change with temperature or gas type? 

**Tip:** Always ensure to identify the type of gas you're working with as it significantly affects the calculations in thermodynamics.

Explore the final pressures of monatomic and diatomic gases undergoing an adiabatic process from V1 to V2. Discover which gas has the lower pressure and the ratio of their pressures expressed in terms of V1 and V2. This analysis is crucial for understanding thermodynamic principles at the high school level.

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