Question

The rms speed of a molecule of oxygen at 127°C
is half that of a molecule of hydrogen at
[NCERT Pg. 330]
(1) 100 K
(2) 273 K
(3) 173 K
(4) 100°C​

Answer 1

Answer:

100 K

Explanation:

TEMPERATURE =273+127= 400 K

$\sqrt{3RT}$/$\sqrt{32}$ =$\sqrt{3RT}$/$\sqrt{2}$

The Vrms of O2 =1/2 of Vrms of H2

Therefore  $\sqrt3RT$/$\sqrt{32}$=$\sqrt{3RT}$/$\sqrt{2}$×1/2

Squaring on both sides

3RT/32=1/4 ×3RT/2

subsituting for T and canceling common factors we get,

                           T=100 K

Answer 2

Answer:

The RMS speed of oxygen at 127°C is half that of a molecule of hydrogen at  100K.i.e. option (1).

Explanation:

The RMS speed is defined as the square root of the mean of the square of speeds of all molecules. The root-mean-square speed takes into account both molecular weight and temperature, two factors that affect a material's kinetic energy. It is given as,

$V_R_M_S=\sqrt{\frac{3RT}{M}}$          (1)

R=universal gas constant

T=temperature of the gas in Kelvin

M=molecular mass of the gas

RMS speed of hydrogen will be given as,

The molecular mass of hydrogen(H₂)=2 g/mol

$V_R_M_S_1=\sqrt{\frac{3RT_1}{2}}$                  (2)

Now the RMS speed of oxygen will be given as,

The molecular mass of oxygen (O₂)=32 g/mol

Temperature=273+127=400K

$V_R_M_S_2=\sqrt{\frac{3R\times 400}{32}}$               (3)

As per the question,

$\sqrt{\frac{3R\times 400}{32}}=\frac{1}{2}\sqrt{\frac{3RT_1}{2}}$

$T_1=100K$

Hence, The RMS speed of a molecule of oxygen at 127°C is half that of a molecule of hydrogen at 100K i.e. option(1).