The r.m.s. velocity of the molecules in the sample of helium is 5/7th that of the molecules in the sample of hydrogen. If the temperature of the hydrogen sample is 0°C that of helium is (a) 0°C (b) 0°K (c) 273°C (d) 100°C

The problem involves understanding the relationship between the root mean square (r.m.s.) velocity of gas molecules and temperature. To tackle this, we can use the formula for r.m.s. velocity, which is given by the equation: v_rms = √(3RT/M) where: - **v_rms** is the root mean square velocity, - **R** is the universal gas constant, - **T** is the absolute temperature in Kelvin, and - **M** is the molar mass of the gas. In this scenario, we know that the r.m.s. velocity of helium is \( \frac{5}{7} \) that of hydrogen, and we are given that the temperature of hydrogen is 0°C. We need to find the temperature of helium.Let's break this down step-by-step:

Step 1: Convert Temperature to Kelvin

The temperature of hydrogen is 0°C. We need to convert this to Kelvin using the conversion formula: T(K) = T(°C) + 273.15 So, \( T_{H2} = 0 + 273.15 = 273.15 \) K.

Step 2: Set Up the Relationship Between Velocities

According to the problem, the r.m.s. velocity of helium (He) can be expressed as: v_{rms, He} = \frac{5}{7} v_{rms, H2} Substituting the r.m.s. formula gives us: √(3R(T_{He}/M_{He})) = \frac{5}{7} √(3R(T_{H2}/M_{H2}))

Step 3: Cancel Out Common Terms

Since R and the square root of 3 appear in both sides of the equation, we can cancel them out, which simplifies our equation to: √(T_{He}/M_{He}) = \frac{5}{7} √(T_{H2}/M_{H2})

Step 4: Substitute Known Values

Now we can substitute \( T_{H2} \) and the molar masses. The molar mass for hydrogen (H2) is approximately 2 g/mol, and for helium (He), it is about 4 g/mol. Thus, we rewrite the equation: √(T_{He}/4) = \frac{5}{7} √(273.15/2)

Step 5: Solve for Helium Temperature

Squaring both sides to eliminate the square roots gives us: (T_{He}/4) = \left(\frac{5}{7}\right)^2 \times \frac{273.15}{2} Calculating the right side: \(\left(\frac{5}{7}\right)^2 = \frac{25}{49}\) Now, we find: T_{He} = 4 \times \left(\frac{25}{49}\right) \times \frac{273.15}{2} This results in: T_{He} = 4 \times \frac{25 \times 273.15}{98} Calculating further yields approximately \( T_{He} \approx 42.06 \) K.

Step 6: Convert Back to Celsius

To convert \( T_{He} \) back into Celsius: T(°C) = T(K) - 273.15 Thus, T_{He} ≈ 42.06 - 273.15 = -231.09°C This value does not match any of the provided options directly. However, it indicates that there's a significant temperature difference between the two gases.

Final Consideration

Among the answer choices presented: (a) 0°C, (b) 0°K, (c) 273°C, (d) 100°C, the correct interpretation considering the calculations and the physical context is that helium would not be at a temperature that matches any of these options directly, but the most logical choice given the context of the problem would suggest lower temperatures as valid. Thus, even though the calculation leads to a temperature not listed, it serves to illustrate the relationship between molecular mass, temperature, and velocity in gases. In practical problems, ensure to verify the ranges of temperatures and conditions under which the gases are compared.