Chapter 20: Q. 43 (page 568)

You are watching a science fiction movie in which the hero shrinks down to the size of an atom and fights villains while jumping from air molecule to air molecule. In one scene, the hero's molecule is about to crash head-on into the molecule on which a villain is riding. The villain's molecule is initially $50$ molecular radii away and, in the movie, it takes $3.5 s$ for the molecules to collide. Estimate the air temperature required for this to be possible. Assume the molecules are nitrogen molecules, each traveling at the rms speed. Is this a plausible temperature for air?

Short Answer

Expert verified

Temperature for the air is $5.8 \times 10^{- 22} K$ ,unrealistically low.

Step by step solution

Formula for root mean square

Root mean square with distance be $d$ covered in time $t$ ,

$v_{rms} = 2 \frac{d}{t} . . . . . . . 1$

We may calculate the rms speed using the formula,

$v_{τms} = \sqrt{\frac{3 k_{B} T}{m}} . . . . . . . . . . . . . . . . 2$

Calculation for temperature

where $m$ is the mass of one particle.

Equating $1$ and $2$ equation,

localid="1648641658334" $2 \frac{d}{t} = \sqrt{\frac{3 k_{B} T}{m}}$

localid="1648641663890" $\frac{3 k_{B} T}{m} = \frac{4 d^{2}}{t^{2}} \Rightarrow T = \frac{4 {md}^{2}}{3 k_{B} t^{2}}$

Substituting localid="1648642043882" $m = \frac{M}{N_{A}}$ andlocalid="1648641689763" $k_{B} N_{A} = R$ we get,

localid="1648641709109" $T = \frac{4 {Md}^{2}}{3 k_{B} N_{A} t^{2}} = \frac{4 {Md}^{2}}{3 {Rt}^{2}}$

localid="1648641723376" $T = \frac{4 (50 r)^{2} M}{3 {Rt}^{2}}$

localid="1648641744912" $T = \frac{4 \cdot {(5 \cdot 10^{- 9})}^{2} \cdot 0.028}{3 \cdot 8.314 \cdot 3 . 5^{2}} = 5.8 \cdot 10^{- 22} K$ ,unrealistically low

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Formula for root mean square

Calculation for temperature

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