Question

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22. At what temperature do the atoms of argon (Ar) gas have the same rms speed as the molecules of oxygen (O2) gas at 25°C? The molar mass of oxygen is 15.9994 g/mol, and the molar mass of argon is 39.948 g/mol. ANSWER: 99ºC

29 A stone is dropped into a well. The sound of the splash is heard 3.5 s later. What is the depth of the well? Take the speed of sound in air to be 343 m/s. ANSWER h = 54.8m

Answer 1

a)

rms speed of any gas is defined as

V_rms = [ 3RT/M]^0.5 -----(1)

where, T = temperature in K

M = molecular mass of gas

R = universal gas constant

now according to question,

V_{rms, O} = V_{rms, Ar}

hence,

[ 3RT_O/M_O ]^0.5 = [ 3RT_Ar/M_Ar]^0.5

or,

T_O/M_O = T_Ar / M_Ar ----(2)

given , T_O= 25+273.15 = 298.15 K

atomic mass of oxygen =15.9994 g/mol

M_O = molecular mass of oxygen = 2*15.9994= 31.9988 g/mol

M_Ar = 39.948 g/mol

put in 2, we get

T_Ar= T_O * M_Ar /M_O

T_Ar= 298.15 * 39.948 / 31.9988 = 372.216 K = 372.216-273.15 = 99.066 C

b)

firstly, the stone is falling under gravity. Let t₁ be the time from from air to well

d =depth and initially stone is at rest , so u=0

hence, from 2nd equation of motion

So d=1/ 2* g*t₁² ---(1)

After hitting the well, the sound travels back up. Using 343m/s for the speed of sound this gives, time taken t₂ .

d=343×t₂ ----(2)

We also know that:

t₁+t₂=3.5 s ---- (3)

equating 1 and 2, we get

343* t₂=1/ 2* g*t₁²

From (3)

t₂=3.5−t₁

So:

343(3.5−t₁)=1/ 2* g*t₁²

1200.5 - 343*t₁ = 0.5*9.8*t₁²

1200.5 -343*t₁ = 4.9*t₁²

4.9*t₁² +343*t₁ - 1200.5 = 0

equating this quadratic eqation , we get

t₁ = 3.34 s

so, t₂ = 3.5 -3.34 = 0.16 s

hence, d = 343 * 0.16 = 54.88 m