Two balloons are inflated with 2.0 mol of Helium (MM = 4 g/mol) and 2.0 mol of A
ID: 949615 • Letter: T
Question
Two balloons are inflated with 2.0 mol of Helium (MM = 4 g/mol) and 2.0 mol of Argon (MM = 40 g/mol) respectively. The external pressure and temperature are 1.00 atm and 298 K.
1) How will the mass of the Argon balloon compare to the mass of the Helium balloon?
A) 10x heavier B) 10x lighter C) same masses D) more info needed E) none of the above
2) How will the volume of the Argon balloon compare to the volume of the Helium balloon?
A) 10x larger B) 10x smaller C) same volumes D) more info needed E) none of the above
3) How will the density of the Argon balloon compare to the density of the Helium balloon?
A) 10x higher B) 10x lower C) same densities D) more info needed E) none of the above
4) How will the volume of the Argon balloon compare to the volume of the Helium balloon, if both balloons are dipped into liquid nitrogen (T = 77 K)?
A) 10x larger B) 10x smaller C) same volumes D) more info needed E) none of the above
5) How will the density of the Argon balloon compare to the density of the Helium balloon, if both balloons are dipped into liquid nitrogen (T = 77 K)?
A) 10x higher B) 10x lower C) same densities D) more info needed E) none of the above
Explanation / Answer
1.) Mass = MM*mol
Mass of Argon balloon = 4 g/mol * 2
= 8 g
Mass of Helium balloon = 40 g/mol * 2
= 80 g
Ans: B
2.) PV = nRT
V = nRT/P
volume of both the balloons will be equal because everything on the RHS is same for both.
Ans: C
3.) Density = M/V
Since, volume is equal. Density will have same relation as mass.
Ans: B
4.) Same hence nothing on RHS changes.
5.) Ans: C