Nitrogen Gas RMS Velocity Calculation

This problem involves calculating the amount of heat transferred to a given mass of nitrogen gas to double the root mean square (RMS) velocity of its molecules. We will use the concepts of ideal gas laws, kinetic theory of gases, and thermodynamics, specifically the first law of thermodynamics for a constant volume process.

Understanding RMS Velocity and Temperature

The root mean square (RMS) velocity of gas molecules is a measure of the speed of the particles in a gas and is directly related to its absolute temperature. The formula for RMS velocity ($v_{rms}$) is given by:

$$v_{rms} = \sqrt{\frac{3RT}{M}}$$

Where:

• $R$ is the universal gas constant.
• $T$ is the absolute temperature of the gas in Kelvin.
• $M$ is the molar mass of the gas.

From this formula, it is clear that the RMS velocity is directly proportional to the square root of the absolute temperature ($v_{rms} \propto \sqrt{T}$). If the RMS velocity is doubled, the square root of the temperature must also be doubled.

Initial Parameters and Conversions

Let's first list the given parameters and convert them to appropriate SI units:

• Mass of nitrogen gas ($m$) = 56 g
• Initial temperature ($T_1$) = $127^\circ C$
• Gas constant ($R$) = 8.3 J/K mole

Convert the initial temperature from Celsius to Kelvin:

$$T_1 = 127^\circ C + 273 = 400 K$$

Nitrogen gas is diatomic ($N_2$). Its molar mass ($M$) can be calculated as:

$$M = 2 \times (\text{atomic mass of Nitrogen}) = 2 \times 14 \, \text{g/mol} = 28 \, \text{g/mol}$$

Now, calculate the number of moles ($n$) of nitrogen gas:

$$n = \frac{\text{mass of gas}}{\text{molar mass}} = \frac{m}{M} = \frac{56 \, \text{g}}{28 \, \text{g/mol}} = 2 \, \text{mol}$$

Calculating New Temperature for Doubled RMS Velocity

If the RMS velocity is doubled, let the new temperature be $T_2$.

$$v_{rms2} = 2 \times v_{rms1}$$

Substituting the RMS velocity formula:

$$\sqrt{\frac{3RT_2}{M}} = 2 \times \sqrt{\frac{3RT_1}{M}}$$

Squaring both sides of the equation:

$$\frac{3RT_2}{M} = 4 \times \frac{3RT_1}{M}$$

Simplifying the equation, we get:

$$T_2 = 4 \times T_1$$

Now, substitute the value of $T_1$:

$$T_2 = 4 \times 400 K = 1600 K$$

The change in temperature ($\Delta T$) is:

$$\Delta T = T_2 - T_1 = 1600 K - 400 K = 1200 K$$

Heat Transferred Calculation (Thermodynamics)

The nitrogen gas is enclosed in a vessel, which implies that the volume of the gas remains constant (isochoric process). For an isochoric process, the work done ($\Delta W$) by the gas is zero.

According to the First Law of Thermodynamics:

$$\Delta Q = \Delta U + \Delta W$$

Since $\Delta W = 0$ for a constant volume process:

$$\Delta Q = \Delta U$$

The change in internal energy ($\Delta U$) for an ideal gas is given by:

$$\Delta U = n C_v \Delta T$$

Where $C_v$ is the molar specific heat capacity at constant volume. For a diatomic gas like nitrogen, the molar specific heat capacity at constant volume ($C_v$) is $\frac{5}{2}R$.

$$C_v = \frac{5}{2}R$$

Substitute this into the equation for $\Delta Q$:

$$\Delta Q = n \left(\frac{5}{2}R\right) \Delta T$$

Now, substitute the values we have calculated:

• $n = 2 \, \text{mol}$
• $R = 8.3 \, \text{J/K mole}$
• $\Delta T = 1200 \, \text{K}$

$$\Delta Q = 2 \, \text{mol} \times \frac{5}{2} \times 8.3 \, \text{J/K mole} \times 1200 \, \text{K}$$

$$\Delta Q = 5 \times 8.3 \times 1200 \, \text{J}$$

$$\Delta Q = 41.5 \times 1200 \, \text{J}$$

$$\Delta Q = 49800 \, \text{J}$$

Converting Joules to Kilojoules (1 kJ = 1000 J):

$$\Delta Q = \frac{49800}{1000} \, \text{kJ} = 49.8 \, \text{kJ}$$

This value is approximately 50 kJ.

Summary of Steps and Result

The problem was solved by following these steps:

1. Converted the initial temperature from Celsius to Kelvin.
2. Calculated the number of moles of nitrogen gas using its mass and molar mass.
3. Used the relationship between RMS velocity and temperature to determine the new temperature required to double the RMS velocity.
4. Calculated the change in temperature.
5. Applied the First Law of Thermodynamics for an isochoric (constant volume) process to find the heat transferred, using the molar specific heat capacity at constant volume for a diatomic gas.

The amount of heat transferred to the nitrogen gas is approximately 50 kJ.