Answer to Question #340927 in Chemistry for shimmy

Question

2. The owner of Molina’s Toy Balloon at Public Market has a 200.0 L Helium gas at a temperature of 280C and a pressure of 760 mmHg and transferred to a tank with a volume of 68.0 L. What is the internal pressure in atm of the tank if the temperature is maintained?

Formula: P1 V1= P2V2 therefore, P2= P1V1 V2 V2 V2

Answer 1

Given:

P₁ = 760 mmHg

V₁ = 200.0 L

T₁ = T₂ = 28°C = constant

P₂ = unknown

V₂ = 68.0 L

Formula: P₁V₁ = P₂V₂

Solution:

Since the temperature and amount of gas remain unchanged, Boyle's law can be used.

Boyle's gas law can be expressed as: P₁V₁ = P₂V₂

To find the internal pressure of the tank, solve the equation for P₂:

P₂ = P₁V₁ / V₂

P₂ = (760 mmHg × 200.0 L) / (68.0 L) = 2235.3 mmHg

Convert mmHg to atm:

P₂ = (2235.3 mmHg) × (1 atm / 760 mmHg) = 2.94 atm

P₂ = 2.94 atm

Answer: The internal pressure of the tank is 2.94 atm