A gas at 430.0K occupies 18.0L of space and has a pressure of 2.40atm. What is the temperature in Celsius if the volume expands to 27.0L and the pressure decreases to 1.10atm

25.5 degrees celsius

Explanation:

Use the combined gas law.

P1V1 / T1=P2V2 / T2

Let the subscript 1 represent the initial 18.0L of gas and the subscript 2 represent the gas at the final volume of 27.0L.

T1=430.0K, P1=2.40atm, V1=18.0L, P2=1.10atm, V2=27.0L, and T2 is unknown.

Substitute the known values into the combined gas law equation.

P1V1 / T1=P2V2 / T2→(2.40atm)(18.0L) / (430.0K)=(1.10atm)(27.0L) / T2

Solve the equation for T2, simplify, and report the answer with three significant figures.

T2=P2V2T1 / P1V1= (1.10atm)(27.0L)(430.0K) / (2.40atm)(18.0L)=295.625K

To get the temperature in Celsius, subtract 273.15.

T2=22.5∘C