A sample of helium had a volume of 3.20x102 mL at STP. What will be it’s new volume of the temperature is increased to 425.0k and it pressure is increased to 3.50 atm?

Combined gas law put together both Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that "the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.

It is expressed as;

P₁V₁/T₁ = P₂V₂/T₂

Given the data in the question;

Initial volume of helium gas V₁ = 3.20 × 10²mL = 0.32L

At standard remperature and pressure

Initial pressure P₁ = 1.0atm

Initial temperature T₁ = 273.15K

Final pressure P₂ = 3.50atm
Final temperature T₂ = 425.0K
Final volume V₂ = ?

To calculate the new volume of the helium gas, we subtsitute our given values into the expression above.

P₁V₁/T₁ = P₂V₂/T₂

P₁V₁T₂ = P₂V₂T₁

V₂ = P₁V₁T₂ / P₂T₁

V₂ = ( 1.0atm × 0.32L × 425.0K ) / ( 3.50atm × 273.15K )

V₂ = 136LatmK / 956.025atmK

V₂ = 0.14L

V₂ = 1.40×10²mL

Therefore, the new volume of the sample of helium gas as its temperature and pressure increases to the given values is 1.40×10²mL.

Learn more about combined gas law here: brainly.com/question/25944795