1)If I have 45 liters of helium in a balloon at 250 C and increase the temperature of the balloon to 550 C, what will the new volume of the balloon be?

V₂ = 70.80 L

Explanation:

Given data:

Initial volume of He = 45 L

Initial temperature = 250 °C

Final temperature = 550 °C

Final volume = ?

Solution:

Initial temperature = 250 °C (250+273.15 K= 523.15 K)

Final temperature = 550 °C (550 + 273.15 k= 823.15 K)

According to Charles Law.,

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 45 L × 823.15  K / 523.15 K

V₂ = 37041.75 L.K / 523.15 K

V₂ = 70.80 L

right

explanation:

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.