Question

The volume of balloon filled with 4.0g of He at 22$^ \circ C$ and 720mm of Hg is:
(A) 25.565 litre
(B) 20 litre
(C) 15 litre
(D) 30 litre

Answer 1

First of all we will calculate the number of moles of helium. Then we will use the ideal gas equation to calculate the volume of the given gas because temperature, pressure and given weight are given in the question.

Formula used:
-Ideal Gas Law: PV = nRT
Where, P = pressure;
V = volume;
n = moles;
R = gas constant;
T = temperature.

Complete step by step answer:
-The given weight of helium in the question is = 4.0g
We know that the molecular weight of helium is = 4g
So, we will now calculate the number of moles of helium using the given formula:
$n = \dfrac{W}{M}$
= $\dfrac{4}{4}$ = 1
So, the number of moles of He are: 1 mole.

-The question gives us the following values:
Temperature (T) = 22$^ \circ C$ = 295K
Pressure (P) = 720mm of Hg = $\dfrac{{720}}{{760}}$ atm
We know that: R = 0.0821$L.atm.mo{l^{ - 1}}.{K^{ - 1}}$

We need to calculate the volume of helium gas. We will do this using the ideal gas equation:
PV = nRT
$\dfrac{{720}}{{760}} \times V = 1 \times 0.0821 \times 295$
V = $\dfrac{{1 \times 0.0821 \times 295 \times 760}}{{720}}$
V = 25.565 L
So, the volume of the balloon filled with helium in the given conditions is 25.565 litre.
So, the correct answer is “Option A”.

Note: In such questions the most common mistake made is the value of R (gas constant). The value of R should be taken according to the units of temperature, pressure and volume.
R = 8.314 $Pa.{m^3}.{K^{ - 1}}.mo{l^{ - 1}}$
R = 0.08314 $bar.L.{K^{ - 1}}.mo{l^{ - 1}}$
R= 8.314 $J.{K^{ - 1}}.mo{l^{ - 1}}$
R= 0.082057 $L.atm.{K^{ - 1}}.mo{l^{ - 1}}$