A sample of gas initially occupies 3.50 L at a pressure of 795 torr at 32.0oC. What will the temperature be if the pressure is changed to 4.00 atm, and the volume is changed to 1.50 L

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 V₁ = 3.50L

Initial pressure P₁ = 795torr = 1.04605atm

Initial temperature T₁ = 32.0°C = 305.15K

Final volume V₂ = 1.50L

Final pressure P₂ = 4.00atm

Final temperature T₂ = ?

To calculate the new temperature the gas, we subtsitute our given values into the expression above.

P₁V₁/T₁ = P₂V₂/T₂

P₁V₁T₂ = P₂V₂T₁

T₂ = P₂V₂T₁/P₁V₁

T₂ = ( 4.00atm × 1.50L × 305.15K ) / ( 1.04605atm × 3.50L )

T₂ = 1830.9LatmK / 3.661175Latm

T₂ = 500.085K

T₂ = 226.9°C

Therefore, as the volume was decreased and pressure increased to the given values, the new temperature of the ideal gas is 226.9°C.

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