Given:
V1 = 4m3
T1 = 290k
P1 = 475 kpa = 475000 Pa
V2 = 6.5m3
T2 = 277K
Required:
P
Solution:
n = PV/RT
n = (475000 Pa)(4m3) / (8.314 Pa-m3/mol-K)(290k)
n = 788 moles
P = nRT/V
P = (788 moles)(8.314 Pa-m3/mol-K)(277K)/(6.5m3)
P = 279,204 Pa or 279 kPa
Explanation:
Since, the given balloon contains equal volume of dry air at the same temperature and pressure therefore, moles of air filled is equal to the moles of helium gas.
Moles of air is 228363.69 moles and molar mass of air is 28.8 g/mol.
As we know that,
No. of moles of air =
228363.69 mol =
mass = 6576874.341 g
Therefore, mass left is as follows.
6576874.341 g - 913454.77 g (as 913454.77 g = mass of He)
= 5664419.57 g
Thus, we can conclude that mass left is 5664419.57 g.
Answer:
Aproperty of matter by which it remains at rest or in motion in the same straight line unless acted upon by some external force