A spherical balloon of volume 4.00 × 103 cm 3 contains helium at a pressure of 1.20 × 105 Pa. How many moles of helium are in the balloon if the average kinetic energy of the helium atoms is 3.60 × 10-22 J ?

Question

A spherical balloon of volume 4.00 × 103 cm 3 contains helium at a pressure of 1.20 × 105 Pa. How many moles of helium are in the balloon if the average kinetic energy of the helium atoms is 3.60 × 10-22 J ? (A) 3.32 mol (B) 2.16 mol (C) 4.12 mol (D) 2.8 mol

Tardigrade Admin · Accepted Answer

The gas temperature must be that implied by We know kinetic energy of a molecule is given by

\frac{1}{2} m_{0} \overset{v}{ˉ}^{2} = \frac{3}{2} k_{B} T

\Rightarrow T = \frac{2}{3} (\frac{\frac{1}{2} m _{0} v ˉ ^{2}}{k _{B}})

\Rightarrow T = \frac{2}{3} (\frac{3.60 \times 1 0 ^{- 22}}{1.38 \times 1 0 ^{- 23} J / K})

= 17.4 K

Now

P V = n RT \Rightarrow n = \frac{P V}{RT}

n = \frac{( 1.20 \times 1 0 ^{5} N / m ^{2} ) ( 4.00 \times 1 0 ^{- 3} m ^{3} )}{( 8.314 j / m o l . K ) ( 17.4 K )} = 3.32 m o l

Q. A spherical balloon of volume 4.00×103cm3 contains helium at a pressure of 1.20×105Pa. How many moles of helium are in the balloon if the average kinetic energy of the helium atoms is 3.60×10−22J?

3.32 mol

2.16 mol

4.12 mol

2.8 mol

The gas temperature must be that implied by We know kinetic energy of a molecule is given by 21​m0​vˉ2=23​kB​T ⇒T=32​(kB​21​m0​vˉ2​) ⇒T=32​(1.38×10−23J/K3.60×10−22​) =17.4K Now PV=nRT⇒n=RTPV​ n=(8.314j/mol.K)(17.4K)(1.20×105N/m2)(4.00×10−3m3)​=3.32mol

Q. A spherical balloon of volume $4.00 \times 1 0^{3} c m^{3}$ contains helium at a pressure of $1.20 \times 1 0^{5} P a$ . How many moles of helium are in the balloon if the average kinetic energy of the helium atoms is $3.60 \times 1 0^{- 22} J ?$