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Question: To what temperature the hydrogen at \(327^\circ \,C\) be cooled at constant pressure, so that the ro...

To what temperature the hydrogen at 327∘ C327^\circ \,C be cooled at constant pressure, so that the root mean square velocity of its molecules become half of its previous value
A. −123∘ C - 123^\circ \,C
B. 23∘ C23^\circ \,C
C. −100∘ C - 100^\circ \,C
D. 0∘ C0^\circ \,C

Explanation

Solution

Here, we will use the concept of the kinetic theory of gases which tells us that the root means square velocity is directly proportional to the square root of the temperature. Now, to calculate the temperature we will divide Vrms1{V_{rms1}} by Vrms2{V_{rms2}}. Here, the root mean square velocity of the molecules of hydrogen will become half of its previous value.

Complete step by step answer:
Now, it is given in the question that the hydrogen is at 327∘ C=327+273=600 K327^\circ \,C = 327+273 = 600\,K. Here, we will change the temperature from Celsius into Kelvin. Also, the root mean square value of the molecules of the hydrogen becomes half of its previous value. Therefore, if we consider Vrms1=V{V_{rms1}} = V
Than Vrms2=V2{V_{rms2}} = \dfrac{V}{2}
Now, according to the kinetic theory of gases, the root mean square velocity of voltage is related to the temperature as
Vrms∝T{V_{rms}} \propto \sqrt T
Now, taking ratios of both the root mean square velocities Vrms1{V_{rms1}} by Vrms2{V_{rms2}} as is given below
Vrms1Vrms2=T1T2\dfrac{{{V_{rms1}}}}{{{V_{rms2}}}} = \dfrac{{\sqrt {{T_1}} }}{{\sqrt {{T_2}} }}
Now, putting the values of Vrms1{V_{rms1}} , Vrms2{V_{rms2}} , T1{T_1} and T2{T_2} in the above equation, we get
VV2=600T2\dfrac{V}{{\dfrac{V}{2}}} = \dfrac{{\sqrt {600} }}{{\sqrt {{T_2}} }}
⇒ T2=6002\Rightarrow \,\sqrt {{T_2}} = \dfrac{{\sqrt {600} }}{2}
⇒ T2=24.492\Rightarrow \,\sqrt {{T_2}} = \dfrac{{24.49}}{2}
⇒ T2=12.24\Rightarrow \,\sqrt {{T_2}} = 12.24
⇒ T2=149.81K\Rightarrow \,{T_2} = 149.81K
⇒ T2≃150K\Rightarrow \,{T_2} \simeq 150K
Now, the above temperature is in kelvin, therefore, we will change the above temperature into Celsius as
T2=(150−273)∘ C{T_2} = \left( {150 - 273} \right)^\circ \,C
∴ T2=−123∘ C\therefore\,{T_2} = - 123^\circ \,C
Therefore, the hydrogen will be cooled at a temperature −123∘ C - 123^\circ \,C such that the root mean square value of its molecules becomes half of its previous value.

Hence, option A is the correct option.

Note: Remember here to change the temperature into Kelvin. We are changing the temperature of hydrogen because the temperature in the root mean square velocity will be in kelvin. Also, after calculating the temperature we will change it to Celsius because the answer required is in Celsius.