A gas that occupies 50.0 liters has its volume increased to 68 liters when the pressure was changed to 3.0 ATM. What was the original pressure of the gas in ATM? In mmHg? In kPa?

4.1 atm = 3,116 mmHg = 415.4 kPa

Explanation:

According to Boyle's law, as volume is increased the pressure of the gas is decreased. That can be expressed as:

P₁ x V₁= P₂ x V₂

Where P₁ and V₁ are the initial pressure and volume respectively, and P₂ and V₂ are final pressure and volume, respectively.

From the problem, we have:

V₁= 50.0 L

V₂= 68.0 L

P₂= 3.0 atm

Thus, we calculate the initial pressure as follows:

P₁= (P₂ x V₂)/V₁= (3.0 atm x 68.0 L)/(50.0 L)= 4.08 atm ≅ 4.1 atm

To transform to mmHg, we know that 1 atm= 760 mmHg:

4.1 atm x 760 mmHg/1 atm = 3,116 mmHg

To transform to kPa we use: 1 atm= 101.325 kPa

4.1 atm x 101.325 kPa = 415.4 kPa

i'm assuming that it's c.) red bank..

the answer would be b i believe