`suppose you have a helium balloon that is holding 2.00 l of he at a pressure of 15.18 psi. if it's temperature started out as 310.15 k, but was heated to 331.15 k, resulting in an expansion of volume to 3.24 l, what is the final pressure in atm?

P = 0.68 atm.

Explanation:

Assuming that we can treat the helium as an ideal gas, we can apply the ideal gas equation to both states, as follows:

Initial State:

P₁*V₁ = n*R*T₁

Final State:

P₂*V₂ = n*R*T₂

Dividing both sides, and arranging terms, we get:

=

Replacing by the givens, we can solve for P₂:

P₂ = 331.15 K * 2.00 L * 15.18 psi / (310.15 K * 3.24 L) = 10.00 psi

As the result is requested in atm, we need to convert from psi (pound/sq in) to atm, as follows:

P(atm) = 10.00 psi* 1/14.696 atm/psi = 0.68 atm

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pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y