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To what temperature should the hydrogen at $ 327 ^\circ C$ cooled at constant pressure, so that the root mean square velocity of its molcules become half of its previous value
$ \nu_{rms} = \sqrt { 3RT \over Mo } \Rightarrow \nu_{rms} \alpha \sqrt T $ $ \therefore T \alpha \nu_{rms} ^2 $
At what temperature is the root mean square velocity of gaseous hydrogen molecules equal to that of oxygen molecules at $ 47 ^\circ C$ ?
$ \nu_{rms} = \sqrt { 3RT \over Mo } \Rightarrow T \alpha M_o $ (\nu_{rms} . R \rightarrow constant ) $
The root mean square velocity of the molecules in a sample of helium is 5/7 th that of the molecules in a sample of hydrogen. If the temperature of hydorgen sample is $0 ^\circ C$ , then the temperature of the helium sample is about
$ \nu_{rms} = \sqrt { 3RT \over Mo } \Rightarrow \nu _{rms} \alpha \sqrt { T \over Mo} $ $ \therefore { \nu_{He} \over \nu_{H_2}} ={ 5 \over 7} = \sqrt { T_{He} \times M_{H_2} \over T_{H2} M_{He} } \Rightarrow = { 25 \over 49 } \times {4 \over 2 } \times 273 $ $ \approx 273 K \approx 0 ^ \circ C$
At room temperature ( $ 27 ^\circ C$ ), the rms speed of the molecules of certain diatomic gas is found to be 1930 m/s. The gas is
$ \nu_{rms} = \sqrt { 3RT \over Mo } \Rightarrow M_o = { 3RT \over \nu_{rms} ^2 } = { 3 \times 8.3 \times 300 \over (1920)^2} = 2 \times 10^{-3} kg = 2g $ Gas is hydrogen
If three molecules have velocities 0.5, 1 and 2 the ratio of rms speed and average speed is (The velocities are in km/s)
rms speed, $ \nu_{rms} = \sqrt { \nu_1^2 +\nu_2^2 + \nu_3^2 \over 3 }$ average speed .$ (\nu) = { \nu_1 +\nu_2 + \nu_3 \over 3 }$
At what temperature pressure remaining constant will the rms speed of a gas molecules increase by 10% of the rms speed at NTP?
$ { ( \nu_{rms} )_t \over ( \nu_{rms})_o} = \sqrt { T \over T_o} and rms speed increases by 10 \% $
$ \therefore 1.1 = \sqrt { 273 + t \over 273} or {273 + t \over 273 } = (1.1)^2 = 1.21 $
$ \therefore t = 273 ( 1.21 - 1 ) = 57.3 C$
When temperature of an ideal gas is increased from $ 27 ^\circ C$ to $ 227 ^\circ C$ , its rms speed changed from $400 ms^{–1}$ to V . The V is
$ \nu_{rms} \alpha \sqrt T $ $ \therefore { \nu_2 \over \nu_1} = \sqrt { T_2 \over T_1} $
The temperature of an ideal gas is increased from $ 27 ^\circ C$ to $ 127 ^\circ C$ , then percentage increase in rms is
$ \nu_{rms} = \sqrt { 3RT \over M_o} $ $ \% increase in \nu_{rms} = { \sqrt {3RT_2 \over M_o} - \sqrt{ 3RT_1 \over M_o } \over \sqrt { 3RT_1 \over M_o} } \times 100 \% = { 20 - 17.32 \over 17.32} \times 100 \% = 15.5 \% $
Let A and B the two gases and given : $ { T_A \over T_B } = 4 { M_A \over M_B } $ Where T is the temperature and M is molecular mass. If $ \nu_A $ and $ \nu_ B $ are the r.m.s speed, then the ratio $ { \nu_A \over \nu_B} $ will be equal to
${T_A \over M_A} = 4 { T_B \over M_B} \Rightarrow \sqrt { T_A \over M_A} = 2 \sqrt { T_B \over M_B}$ $ \Rightarrow \sqrt { 3RT_A \over M_A} = 2 \sqrt { 3RT_B \over M_B} \Rightarrow \nu_A = 2 \nu_B \Rightarrow { \nu_A \over \nu_B} = 2 $
The rms. speed of the molecules of a gas in a vessel is $ 400 ms^{-1}$ . If half of the gas leaks out, at constant temperature, the r.m.s speed of the remaining molecules will be
Since temperature is constant So $ \nu_{rms} $ remains same
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