A high altitude balloon contains 6.0 g of helium in 10'Lat 240 K. Assuming ideal gas behaviour, how
many grams of hellum would have to be added to increase the pressure to 4,0 x 10 atm?
(A) 1
(B) 1.2
(C) 1.5
(D) 2.0
Answers
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0
Answer:
P
1
=
6.81
4
×
0.0821×250
1.1×104
P1=6.814×0.0821×2501.1×104
=3.012×
10
−3
=3.012×10-3
P
f
=4×
10
−3
Pf=4×10-3
△P
↑
⏐
=1×
10
−3
△P↑=1×10-3
n
He-added
=
10
−3
×1.16×
10
4
0.0821×250
nHe-added=10-3×1.16×1040.0821×250
wt=
n
He
×4=2.24gm.
Explanation:
Brainy Lady
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