Helium in a steel tank is at 250 kPa, 300 K with a volume of 0.1m3. It is used to fill a balloon of helium.

When thetank pressure drops to 150 kPa, the flow of helium stops by itself.

If all the helium still is at pressure in the balloon varieslinearly with volume from 100 kPa (V=0) to the final 150 kPa.

How much heat transfer took place?

Our answer is Q = 13.3kj

Explanation:

We are given the following data for helium in a steel tank that is used to filla balloon:

T₁ = 300k

T₂ = 300k

P₁ = 250kPa

p₂=150kPa

V₁ = 0.1m³

From ideal gas ESO we can calculate stage 2 volume

P₁. V₁ = mR. T₁

P₂. V₂ = mR. T₂

Stage 1 and 2 temperature are the same, so final volume is equal to:

V₂ = V₁. P₁ / P₂ = 0.1. 250/150 = 0.16667m³

Ballon volume is equal to:

vB = V₂ _V₁ = 0.16667 - 0.1 = 0.016667m³

Work done in process is equal to:

W = ∫ PdV = 1/2. (250 + 150).0.06667

=13.3kj

Calculte the heat transfer:

Q-W = m (u₂ - u₁) = m. Cv (T₂ - T₁) = 0

Q=W

Q = 13.3kj