Question

A mixture of helium and neon gases at 319 K contains three times the number of helium atoms as neon atoms. The concentration (n/V) of the mixture is found to be 0.150 mol/L. Assuming ideal behavior, what is the partial pressure of helium?

Answer 1

Answer:

Step 1: Explanation:

We know the ideal gas equation, PV=nRT

where, R=universal gas constant,T=temperature ,P=pressure,V=volume,n=moles.

Step 2: Calculation of partial pressure of helium

Here given "Three times the number of helium atoms as neon atoms" which means

3 He for every 1 Ne means that 3/4 of the atoms in the mixture are He. That means that 3/4 of the moles in the mixture are He atoms.

We know,molarity = n/v where, n=mole and v= volume

hence the ideal equation can act like

P = (n/v)RT since n/v =0.150 mol/ L and T=319 K and R=8.314 L-kPa K^-1mol^-1

The partial pressure will 3/4 times times of total molecules

P_[He] = 3/4 (n/v)RT

P_[He] = 3/4 × 0150 mol/L ×8.314 L-kPa K^-1mol^-1 × 319 K  

P_[He] = 298.4 kPa

Hence, the partial pressure of helium is 298.4 kPa