A sample of gas in a balloon has an initial temperature of 30. ∘C and a volume of 1.62×103 L If the temperature changes to 65 ∘C ?

Part A:
Initial: T₁ = (273 + 30) K = 303 K, V₁ = 1.62 × 10³ L
New: T₂ = (273 + 65) K = 338 K, V₂ = ? L

For a fixed amount of gas at constant pressure: V₁/T₁ = V₂/T₂
Then, V₂ = V₁ × (T₂/T₁)

New volume, V₂ = (1.62 × 10³) × (338/303) L = 1.81 × 10³ L

====
Part B:
Initial: T₁ = (273 + 30) K = 303 K, V₁ = 1.62 × 10³ L
New: T₂ = ? K, V₂ = 3.24 × 10³ L

For a fixed amount of gas at constant pressure: V₁/T₁ = V₂/T₂
Then, T₂ = T₁ × (V₂/V₁)

New temperature, T₂ = 303 × [(3.24 × 10³) / (1.62 × 10³)] K = 606 K = 333 °C

Use the relationship:
V1/T1 = V2/T2

Convert temperatures to Kelvin, plug in and solve for V2 and then in the second problem for T2

Part A :

Initial temp = 30°C = 273 + 30 = 303 K
Initial volume = 1.62*103 L
Final temp = 65°C = 273 + 65 = 338 K
V2 = T2*V1/T1 = 338*1.62*103/303 
V2 = 56398.68/303
V2 = 186.1342574 liters
Therefore, the new volume is 186.13 Liters.

Part B :
We know, the volume is proportional to Temperature
First convert temperature to SI unit, 30°C = 273 + 30 = 303 kelvin

V1/T1= V2/T2
T2 = V2*T1/V1
T2= 3.24*103*303/(1.62*103)
T2 = 606 k
Convert Temp back to °C
606 k = 606-273 = 333°C
Therefore, temperature is 333°C

Mark me favorite if this helps!😊