The kinetic energy of - N - molecules of H 2 is 3 J at -73- C- The kinetic energy of the same sample of H 2 at -127- C is -A- 12 JB- 2-19 JC- 9 JD- 3 J

The correct option is B 2.19 J KE_1/KE_2 = n_1T_1/n_2T_2 T_1 = 273 73 = 200 K T_2 = 273 127 = 146 K 3/KE_2 = N × 200/N × 146 KE_2 = 2.1 ...

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Standard VIII

Chemistry

Kinetic Energy

The kinetic e...

Question

The kinetic energy of $^{'} N^{'}$ molecules of $H_{2}$ is $3 J$ at $- 73^{\circ} C$ . The kinetic energy of the same sample of $H_{2}$ at $- 127^{\circ} C$ is ___.

12 J

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$2.19 J$

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$9 J$

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$3 J$

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Solution

The correct option is B
$2.19 J$

$\frac{K E_{1}}{K E_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}}$

$T_{1} = 273 - 73 = 200 K$ $T_{2} = 273 - 127 = 146 K$

$\frac{3}{K E_{2}} = \frac{N \times 200}{N \times 146}$

$K E_{2} = 2.19 J$

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The kinetic energy of ′N′ molecules of H2 is 3 J at −73∘C. The kinetic energy of the same sample of H2 at −127∘C is ___.

The correct option is B 2.19 J KE1KE2 = n1T1n2T2 T1 = 273 − 73 = 200 K T2 = 273 − 127 = 146 K 3KE2 = N × 200N × 146 KE2 = 2.19 J

The kinetic energy of $^{'} N^{'}$ molecules of $H_{2}$ is $3 J$ at $- 73^{\circ} C$ . The kinetic energy of the same sample of $H_{2}$ at $- 127^{\circ} C$ is ___.

The correct option is B
$2.19 J$

$\frac{K E_{1}}{K E_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}}$

$T_{1} = 273 - 73 = 200 K$ $T_{2} = 273 - 127 = 146 K$

$\frac{3}{K E_{2}} = \frac{N \times 200}{N \times 146}$

$K E_{2} = 2.19 J$