Question

5.6 litre of helium gas at stp is adiabatically compressed to 0.7 litre. Assuming that the initial temperature is t1, the work done in the process is

Answer 1

Given,

Initial Volume of the gas(V₁) = 5.6 liter = 5.6 × 10⁻³ m³

Final Volume of the gas(V₂) = 0.7 litre = 0.7 × 10⁻³ m³

Initial Pressure(P₁) = 1 atm = 1.013 × 10⁵ Pa.

Using the Formula,

Work done in adiabatic compression = nR(T₂ - T₁)/(1 - γ)

= (P₂V₂ - P₁V₁)/(1 - γ)

Where, γ is the Ratio of Cp/Cv. or 1 + 2/f

For helium gas, it is mono-atomic, therefore, Degree of Freedom = 3

∴ γ = 1 + 2/f

γ = 1 + 2/3

γ = 5/3

∴ γ = 1.67

Now, we need to find P₂, then we can calculate the work done easily.

Using the Formula,

P₁V₁^γ = P₂V₂^γ

1.012 × 10⁵ × (5.6 × 10⁻³)⁵⁾³ = P₂ ×  (0.7 × 10⁻³)⁵⁾³

P₂ = 1.012 × 10⁵ × (5.6 × 10⁻³)⁵⁾³/ (0.7 × 10⁻³)⁵⁾³

∴ P₂ = 1.012 × 10⁵ × (8 × 10⁻³)⁵⁾³

∴ P₂ =  1.012 × 10⁵ × 2⁵ × 10⁻⁵

P₂ =  1.012 × 32

∴ P₂ = 32.384 Pa.

Now, Work Done can be Calculated from this data.

∴ Work done = (32.384 × 0.7 × 10⁻³ -  1.013 × 10⁵ ×  5.6 × 10⁻³)/(1 - 1.67)

∴ Work done = (0.0226 - 567.28)/-0.67

∴ Work done = + 846.65 Joule.

The Sign of this work done will vary in Physics and in Chemistry. In Physics, It will be negative, whereas In Chemistry it will be positive.

Hope it helps.

Answer 2

Explanation:

5.6 litre of helium gas at stp is adiabatically compressed to 0.7 litre .

• Now, we need to find P₂, then we can calculate the work done easily.

∴ P₂ = 32.384 Pa.

Now, Work Done can be Calculated from this data. ∴ Work done = + 846.65 Joule.