Question

A spherical balloon of volume $4.00 \times 10^{3}\, cm ^{3}$ contains helium at a pressure of $1.20 \times 10^{5} Pa$. How many moles of helium are in the balloon if the average kinetic energy of the helium atoms is $3.60 \times 10^{-22} J ?$

• A. 3.32 mol
• B. 2.16 mol
• C. 4.12 mol
• D. 2.8 mol

Answer 1

Answer: (A)

The gas temperature must be that implied by We know kinetic energy of a molecule is given by
$\frac{1}{2} m_{0} \bar{v}^{2}=\frac{3}{2} k_{B} T$
$\Rightarrow T=\frac{2}{3}\left(\frac{\frac{1}{2} m_{0} \bar{v}^{2}}{k_{B}}\right)$
$\Rightarrow T =\frac{2}{3}\left(\frac{3.60 \times 10^{-22}}{1.38 \times 10^{-23} J / K }\right)$
$=17.4 K$
Now
$P V=n R T \Rightarrow n=\frac{P V}{R T}$
$n=\frac{\left(1.20 \times 10^{5} N / m ^{2}\right)\left(4.00 \times 10^{-3} m ^{3}\right)}{(8.314\, j / mol . K )(17.4\, K )}=3.32\, mol$