When $300 \text{ J}$ of heat given to an ideal gas with $C_p = \frac{7}{2} R$ its temperature raises from $20^\circ\text{C}$ to $50^\circ\text{C}$ keeping its volume constant. The mass of the gas is (approximately) ______

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When $300 \text{ J}$ of heat given to an ideal gas with $C_p = \frac{7}{2} R$ its temperature raises from $20^\circ\text{C}$ to $50^\circ\text{C}$ keeping its volume constant. The mass of the gas is (approximately) _______ g. ($R = 8.314 \text{ J/mol.K}$)

Problem Analysis

We are given the following information for an ideal gas:

Heat supplied, $Q = 300 \text{ J}$
Specific heat at constant pressure, $C_p = \frac{7}{2} R$
Initial temperature, $T_1 = 20^\circ\text{C}$
Final temperature, $T_2 = 50^\circ\text{C}$
Process: Constant volume (isochoric)
Universal gas constant, $R = 8.314 \text{ J/mol.K}$

We need to find the approximate mass ($m$) of the gas in grams.

Calculating Specific Heat at Constant Volume ($C_v$)

For an ideal gas, the relationship between specific heats is given by Mayer's relation: $C_p - C_v = R$

Using the given $C_p = \frac{7}{2} R$: $\frac{7}{2} R - C_v = R$ $C_v = \frac{7}{2} R - R = \frac{5}{2} R$

Substituting the value of $R$: $C_v = \frac{5}{2} \times 8.314 \text{ J/mol.K} \approx 20.785 \text{ J/mol.K}$

Calculating Temperature Change ($\Delta T$)

The change in temperature is: $\Delta T = T_2 - T_1 = 50^\circ\text{C} - 20^\circ\text{C} = 30^\circ\text{C}$

Since the difference is the same in Celsius and Kelvin, $\Delta T = 30 \text{ K}$.

Calculating Number of Moles ($n$)

For a constant volume (isochoric) process, the heat absorbed is given by: $Q = n C_v \Delta T$

Substituting the known values: $300 \text{ J} = n \times (20.785 \text{ J/mol.K}) \times (30 \text{ K})$ $300 = n \times 623.55$ $n = \frac{300}{623.55} \approx 0.481 \text{ mol}$

Calculating Gas Mass ($m$)

The value $C_p = \frac{7}{2} R$ is characteristic of a diatomic ideal gas (considering translational and rotational modes). Common diatomic gases like Nitrogen ($N_2$) have a molar mass ($M$) of approximately $28$ g/mol.

The mass of the gas is calculated using the number of moles ($n$) and the molar mass ($M$): $m = n \times M$

Assuming the gas is diatomic with $M \approx 28$ g/mol: $m \approx 0.481 \text{ mol} \times 28 \text{ g/mol}$ $m \approx 13.47 \text{ g}$

This value is approximately 13 g, consistent with the provided answer range.

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