A gas balloon has a volume of 106.0 liters when the temperature is 45.0 °C and the pressure is 5atm. What will its volume be at 20.0 °C and standard pressure?

Question

1. My first question is, "A gas balloon has a volume of 106.0 liters when the temperature is 45.0 °C and the pressure is 5atm. What will its volume be at 20.0 °C and standard pressure?"
2. My second question is, "A tire has a pressure of 1500 Kpa, and a volume of 2000L at standard temperature. What is the new pressure if the volume decreased to 1000L and temperature was 150K?"

Please help, I don't understand them.

Answer 1

1) From ideal gas law, P₂V₂/P₁V₁ = T₂/T₁ and V₂ = V₁ T₂ P₁ /T₁P₂ = (106.0 L) (20 +273 K) (5 atm) / [(45 +273K) (1 atm)] = 488.3 L.

2) Likewise P₂ = P₁ T₂ V₁ /T₁V₂ = (1500 kPa) (150 K) (2000 L) / [(273 K) (1000 L)] = 1648 kPa.