Heat Supplied to Diatomic Gas with Rotational Modes

To find the amount of heat supplied to the diatomic gas, we apply the First Law of Thermodynamics:

$Q = \Delta U + W$

1. Calculation of Internal Energy Change ($\Delta U$)

A diatomic gas at room temperature possesses 3 translational and 2 rotational degrees of freedom (with vibrational modes inactive). Thus, the total active degrees of freedom is: $f = 3 + 2 = 5$

The molar heat capacity at constant volume ($C_v$) is given by: $C_v = \frac{f}{2}R = \frac{5}{2}R$

Given:

• Number of moles, $n = 1\text{ mol}$
• Universal gas constant, $R = 8.3\text{ J/mol}\cdot\text{K}$
• Temperature rise, $\Delta T = 1.2^\circ\text{C} = 1.2\text{ K}$

The change in internal energy ($\Delta U$) is: $\Delta U = n C_v \Delta T = 1 \times \left(\frac{5}{2} \times 8.3\right) \times 1.2$ $\Delta U = 2.5 \times 8.3 \times 1.2 = 24.9\text{ J}$

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2. Calculation of Work Done ($W$)

The work done by the gas as the piston moves slowly against atmospheric pressure is: $W = P_{atm} \Delta V = P_{atm} \cdot (A \cdot \Delta x)$

Given:

• Atmospheric pressure, $P_{atm} = 100\text{ kPa} = 10^5\text{ Pa}$
• Area of cross-section, $A = 4\text{ cm}^2 = 4 \times 10^{-4}\text{ m}^2$
• Displacement of the piston, $\Delta x = 25\text{ mm} = 25 \times 10^{-3}\text{ m}$

Substituting the values: $W = 10^5 \times (4 \times 10^{-4}) \times (25 \times 10^{-3}) = 1\text{ J}$

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3. Conclusion

The calculated value for the change in internal energy $\Delta U = 24.9\text{ J}$ lies directly between 24.8 J (Option A) and 25 J (Option B):

• Rounding $24.9\text{ J}$ yields $25\text{ J}$.
• Considering standard computational variances in $R$, $24.8\text{ J}$ is also extremely close.

Thus, both Options A and B are correct.