If a 4.50 L container of helium is held under 100 kPa and a larger container (7.0 L), the new pressure is mathematically given as
P=67.28kpa
Question Parameter(s):
At a constant temperature, a 4.50 L container of helium is held under 100 kPa.
The helium is transferred to a larger container (7.0 L),
Generally, the equation for the Ideal gas is mathematically given as
V1P1=V2P2
Therefore
4.50*100=7.0*P2
P=450/7
P=67.28kpa
In conclusion, the new pressure is
P=67.28kpa
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