The temperature at which the rms velocity of SO2(g) is same as that of O2(g) at 320 K, is
Question
The temperature at which the rms velocity of SO2(g) is same as that of O2(g) at 320 K, is
Solution
The root mean square (rms) velocity of a gas is given by the formula:
v_rms = sqrt(3kT/m)
where:
- k is the Boltzmann constant,
- T is the temperature in Kelvin,
- m is the molar mass of the gas.
We are given that the rms velocity of SO2(g) is the same as that of O2(g) at 320 K. Therefore, we can set up the following equation:
sqrt(3kT_SO2/m_SO2) = sqrt(3kT_O2/m_O2)
We can simplify this equation by squaring both sides and cancelling out the common factors:
3kT_SO2/m_SO2 = 3kT_O2/m_O2
This simplifies to:
T_SO2/m_SO2 = T_O2/m_O2
We can rearrange this equation to solve for T_SO2:
T_SO2 = T_O2 * (m_SO2/m_O2)
Substituting the given values:
T_SO2 = 320 K * (64 g/mol / 32 g/mol) = 640 K
Therefore, the temperature at which the rms velocity of SO2(g) is the same as that of O2(g) at 320 K is 640 K.
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