To an evacuated vessel with movable piston under
external pressure of 1 atm, 0.1 mol of He and 1.0
mol of an unknown compound (vapour pressure
0.68 atm.) at 0°(C) are introduced. Considering the
ideal gas behaviour, (the volume in litre) of the
gases at 0°C is close to
Answers
Hi,
Answer: 7 litre
Explanation:
The total external pressure of the vessel, P = 1 atm
The temperature of the vessel = 0℃ = 273 K
Vapour pressure of the unknown compound, P2 = 0.68 atm
∴ Pressure of He, P1 = 1 – 0.68 = 0.32 atm
No. of moles of He present in the vessel, n1 = 0.1 mol
Let the no. of moles of the unknown compound be denoted as “n2” and the total volume of gases in the vessel be denoted as “V” litres.
Using Avogadro’s Law, we have
P1/n1 = P2/n2
⇒ 0.32/0.1 = 0.68/n2
⇒ n2 = 0.68/3.2 = 0.2125 mol
Now, by using the Ideal Gas Law, the total volume of the vessel can be calculated as follows,
PV = nRT
⇒ P * V = (n1+n2) * R * T
Substituting the values,
1 * V = (0.1 + 0.2125) * 0.0821 * 273 ….. [∵ Ideal gas constant R = 0.821 L atm K⁻¹ mol⁻¹]
⇒ V = 0.3125 * 0.0821 * 273 = 7.004 L ≈7 litre
Thus, the total volume of the gases at 0℃ is close to 7 litre.
Hope this helps!!!!