Question

(a) calculate the total and average kinetic energy of 32g methane molecules at 27°c .

(B) cal. the root mean square speed, average speed, and most probable speed of methane at 27°c.​

Answer 1

$\huge{\underline{\underline{\mathbf{Answer:-}}}}$

(A) $3741.3j \: mo {l}^{ - 1}$

(B) Average speed - $629.9 \: m {s}^{ - 1}$

Probable speed - $558.1 \: m {s}^{ - 1}$

$\huge{\underline{\underline{\mathbf{Explanation:-}}}}$

(A) Total K.E. =

$\frac{3}{2} nrt$

✏ $\frac{3}{2} \times \frac{32}{16} mol \times 8.314j {k}^{ - 1} \: mo {l}^{ - 1} \times 300k$

( molar mass of CH4 = 16g mol)

★ $\boxed{7482.6j}$

Average K. E. = $\frac{3}{2} kt$

✏ $\frac{3}{2} \times \frac{8.314}{6.02 \times {10}^{23} } \times 300k$

★ 6.21×10²¹ ★

_________________________________________

(B) Root mean square speed,

$c = \sqrt{ \frac{3rt}{m} }$

✡$\bold{by\:using\: CGS\: unit\:}$

then,

T => 27 + 273 = 300 k

✡ M (for CH4) = 16g

we get,

✏ $u_{c.m.s} = \sqrt{ \frac{3 \times 8.314 \times {10}^{7} \times 300 }{16} }$

★ 6.839×10⁴ cm s ★

✡

$\bold{by\:using\: SI \: unit\:both\:values\:are\:same}$

average speed = 629.9 ms¹

probable speed = 558.1 ms¹

_________________________________________

Answer 2

$\huge{\mathfrak{\red{\underline{\underline{Solution\:1:-}}}}}$

$\star{\bf{\red{\underline{Average\:Kinetic\: energy:-}}}}$

$\sf{K.E.=\frac{3}{2} K_{b}T}$

$\sf{=\frac{3}{2} \frac{RT}{N_{A}}}$

$\sf{Where,}$

$\sf{R=8.314\:JK^{-1}\: mol^{-1}}$

$\sf{N_{A}=6.022 × 10^{23}}$

$\sf{T=27+273=300\:K}$

$\sf{Ē_{kinetic}=\frac{3}{2}×\frac{8.314×300}{6.022×10^{23}}}$

$\sf{=6.21×10^{-21}J}$

${\boxed{\boxed{\bf{So,\:Average \:Kenetic\: energy=6.21×10^{-21}J}}}}$

$\star{\bf{\red{\underline{Total\:Kinetic\: energy:-}}}}$

$\sf{Total\:K.E.=\frac{3}{2}nRT}$

$\sf{=\frac{3}{2}×\frac{32}{16}×8.314×300}$

$\sf{Total\:K.E.=7482.6J}$

${\boxed{\boxed{\bf{So,\:Total\:Kenetic\: energy=7482.6J}}}}$

$\huge{\mathfrak{\red{\underline{\underline{Solution\:2:-}}}}}$

$\sf{We\:have,}$

$\sf{R.M.S., \mu =\frac {3RT}{M}^{\frac{1}{2}}}$

$\sf{Where,}$

$\sf{M=molar\:mass}$

$\sf{In\:case\:of\: oxygen\:gas,}$

$\sf{T=27+273=300K}$

$\sf{R=8.314{(Kg\:m^{2}/{s^{2}})/K\:mol}}$

$\sf{Since\:molar\:mass\:of\: oxygen\:gas=32\:gm/mol}$

$\sf{=3.2×10^{-2}\:Kg/mol}$

$\sf{Hence,}$

$\sf{\mu =\frac {3RT}{M}^{\frac{1}{2}}}$

$\sf{=\frac{3×8.314×300}{3.2×10^{-2}}}^{\frac{1}{2}}$

$\sf{=233831.25^{\frac{1}{2}}}$

$\sf{=483.56\:m/s}$

Hence the R.M speed from oxygen gas at 0°C (273k) = 483.5 m/s.