Calculate the rms speed of Br2 molecules at 23 °C and 1.00 atm. What is the rms speed of Br2 at 23 °C and 1.50 atm?
rms = sqrt(3RT/M)
To calculate the root mean square (rms) speed of gas molecules, we can use the following formula:
v_rms = √((3 * R * T) / M)
where:
- v_rms is the rms speed of the gas molecules
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (23 °C = 23 + 273 = 296 K)
- M is the molar mass of the gas molecule (Br2 = 2 * Atomic mass of Br = 2 * 79.904 g/mol = 159.808 g/mol)
Using this formula, we can calculate the rms speed of Br2 molecules at 23 °C and 1.00 atm:
v_rms = √((3 * 8.314 J/(mol·K) * 296 K) / 159.808 g/mol)
= √(7448.968 J/(mol·K·g/mol))
= √(7448.968 J/(K·g))
≈ √(74.48968 m^2/s^2)
≈ 8.63 m/s
Therefore, the rms speed of Br2 molecules at 23 °C and 1.00 atm is approximately 8.63 m/s.
To calculate the rms speed of Br2 molecules at 23 °C and 1.50 atm, we can use the same formula and the given pressure:
v_rms = √((3 * 8.314 J/(mol·K) * 296 K) / 159.808 g/mol) * (P2 / P1)
= 8.63 m/s * (1.50 atm / 1.00 atm)
= 8.63 m/s * 1.50
= 12.95 m/s
Therefore, the rms speed of Br2 molecules at 23 °C and 1.50 atm is approximately 12.95 m/s.