container A has a sample of monoatomic ideal gas,while container B has sample of diatomic ideal gas. the 2 gases start with the same initial temp. ,volm.,prr., and no. of moles of molecules. each gas now undergoes a reversible adiabatic process that cuts the pressire in half. which sample of gas is now at the higher temp.

To determine which gas sample is at a higher temperature after undergoing a reversible adiabatic process that halves the pressure, we need to consider the properties of monoatomic and diatomic ideal gases, particularly their specific heat capacities.

Understanding Adiabatic Processes

In an adiabatic process, there is no heat exchange with the surroundings. For an ideal gas, the relationship between pressure (P), volume (V), and temperature (T) during an adiabatic process can be described by the equation:

• P1V1^γ = P2V2^γ
• T1V1^(γ-1) = T2V2^(γ-1)

Here, γ (gamma) is the heat capacity ratio, defined as C_p/C_v, where C_p is the specific heat at constant pressure and C_v is the specific heat at constant volume.

Specific Heat Capacities

For monoatomic ideal gases, such as helium or neon, the value of γ is approximately 5/3 (or 1.67). For diatomic ideal gases, like nitrogen or oxygen, γ is about 7/5 (or 1.4). This difference in γ is crucial for our analysis.

Applying the Adiabatic Condition

Since both gases start with the same initial conditions (temperature, volume, pressure, and number of moles), we can analyze how the temperature changes when the pressure is halved.

Calculating Temperature Change

Using the adiabatic relation for temperature and pressure:

• T2 = T1 * (P2/P1)^(γ-1)

Given that the pressure is halved (P2 = P1/2), we can substitute this into the equation:

• T2 = T1 * (1/2)^(γ-1)

Evaluating for Each Gas

Now, let's calculate the final temperature for both gases:

• For the monoatomic gas:
• T2_mono = T1 * (1/2)^(5/3 - 1) = T1 * (1/2)^(2/3)
• For the diatomic gas:
• T2_di = T1 * (1/2)^(7/5 - 1) = T1 * (1/2)^(2/5)

Comparing Final Temperatures

Now we can compare the two final temperatures:

• Since (1/2)^(2/3) is less than (1/2)^(2/5), it follows that:
• T2_mono < T2_di

This means that the final temperature of the monoatomic gas is lower than that of the diatomic gas after the adiabatic process.

Conclusion

In summary, after both gases undergo the reversible adiabatic process that halves their pressure, the diatomic gas will be at a higher temperature than the monoatomic gas. This outcome is primarily due to the differences in their heat capacities, which affect how each gas responds to changes in pressure during an adiabatic process.