The rms speed of a sample of gas is increased by 8% .
What is the percent change in the temperature of the gas?
What is the percent change in the pressure of the gas, assuming its volume is held constant?
PLEASE HELP
To find the percent change in temperature and pressure, we can use the formula for the root mean square (rms) speed of a gas particle:
\(v_{rms} = \sqrt{\frac{3kT}{m}}\)
Where:
- \(v_{rms}\) is the rms speed of the gas particles
- \(k\) is Boltzmann's constant (1.38 x 10^(-23) J/K)
- \(T\) is the temperature of the gas in Kelvin
- \(m\) is the mass of the gas particles
Let's assume the initial values are \(v_{rms_0}\), \(T_0\), and \(P_0\) for the rms speed, temperature, and pressure, respectively. And let's denote the final values as \(v_{rms_1}\), \(T_1\), and \(P_1\).
According to the problem statement, the rms speed increases by 8%. Therefore:
\(v_{rms_1} = (1 + 0.08) \cdot v_{rms_0}\)
To find the percent change in temperature, we need to solve for \(T_1\). Rearranging the formula for \(v_{rms}\), we get:
\(T = \frac{m \cdot v_{rms}^2}{3k}\)
Substituting the initial and final values, we have:
\(T_1 = \frac{m \cdot v_{rms_1}^2}{3k}\)
To find the percent change in pressure, assuming the volume of the gas is held constant, we can use Charles's law, which states:
\(P_1 \cdot T_1 = P_0 \cdot T_0\)
Solving for \(P_1\), we get:
\(P_1 = \frac{P_0 \cdot T_0}{T_1}\)
Now, we can calculate the percent change in both temperature and pressure:
Percent change in temperature:
\(\text{percent change in } T = \left(\frac{T_1 - T_0}{T_0}\right) \cdot 100\)
Percent change in pressure:
\(\text{percent change in } P = \left(\frac{P_1 - P_0}{P_0}\right) \cdot 100\)
By substituting the calculated values of \(T_1\) and \(P_1\) into the respective formulas, we can find the percent changes in temperature and pressure.