Question

to answer this question, first click here to view the associated video. if the temperature of an ideal gas doubles, what happens to the root mean square speed? decreases by a factor of √2. increases by a factor of 4. increases a factor of 2. increases by a factor of √2. what is the internal energy of two moles of a monatomic ideal gas at a temperature of 127°c? 1580 j 9970 j 4990 j 3170 j

Explanation:

First Question (Root Mean Square Speed of Ideal Gas)

Step1: Recall the formula for root mean square (rms) speed of an ideal gas.

The formula for the rms speed \( v_{\text{rms}} \) of an ideal gas is \( v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \) (or equivalently \( v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \)), where \( T \) is the absolute temperature, \( k \) is Boltzmann's constant, \( m \) is the mass of a molecule, \( R \) is the gas constant, and \( M \) is the molar mass.

Step2: Analyze the effect of doubling the temperature.

Let the initial temperature be \( T_1 \) and the final temperature be \( T_2 = 2T_1 \). The initial rms speed is \( v_{1} = \sqrt{\frac{3kT_1}{m}} \), and the final rms speed is \( v_{2} = \sqrt{\frac{3kT_2}{m}} = \sqrt{\frac{3k(2T_1)}{m}} \).

Step3: Find the ratio of the final to initial rms speed.

\( \frac{v_2}{v_1} = \frac{\sqrt{\frac{3k(2T_1)}{m}}}{\sqrt{\frac{3kT_1}{m}}} = \sqrt{\frac{2T_1}{T_1}} = \sqrt{2} \). So the rms speed increases by a factor of \( \sqrt{2} \).

Step1: Recall the formula for internal energy of a monatomic ideal gas.

The internal energy \( U \) of a monatomic ideal gas is given by \( U = \frac{3}{2}nRT \), where \( n \) is the number of moles, \( R = 8.314 \, \text{J/(mol·K)} \), and \( T \) is the absolute temperature.

Step2: Convert the temperature to Kelvin.

The given temperature is \( 127^\circ \text{C} \). To convert to Kelvin, use \( T = 127 + 273 = 400 \, \text{K} \).

Step3: Substitute the values into the formula.

We have \( n = 2 \, \text{mol} \), \( R = 8.314 \, \text{J/(mol·K)} \), and \( T = 400 \, \text{K} \). Plugging into \( U = \frac{3}{2}nRT \):

\( U = \frac{3}{2} \times 2 \times 8.314 \times 400 \)

First, \( \frac{3}{2} \times 2 = 3 \). Then, \( 3 \times 8.314 \times 400 = 3 \times 3325.6 = 9976.8 \, \text{J} \), which is approximately \( 9970 \, \text{J} \).

Answer:

Increases by a factor of \( \sqrt{2} \) (the last option among the first question's choices).

Second Question (Internal Energy of Monatomic Ideal Gas)