A balloon has a volume of 3 L at a temperature of 298 K and a 1 atm. What will the temperature be if the volume is 1.5 L and the pressure is increased to 4 atm?

Respuesta :

Answer: 596 K

Explanation:

[tex]\frac{1 atm * 3L}{298K} =\frac{4 atm * 1.5 L}{x}[/tex]

cross multiply

[tex]1 atm * 3 L * x = 4 atm * 1.5 L * 298 K[/tex]

divide by 1 atm and 3 L to isolate the variable

[tex]x =\frac{4 atm * 1.5 L * 298 K}{1 atm * 3 L}[/tex]

cancel out units

[tex]x =\frac{4 * 1.5 * 298 K}{1 * 3 }[/tex]

multiply numerator and denominator

[tex]x =\frac{1788 K}{3 }[/tex]

divide

[tex]x = 596 K[/tex]