A sample of monoatomic gas is taken at initial pressure of 75 kPa. The volume of the gas is then compressed from 1200 cm³ to 150 cm³ adiabatically. In this process, the value of workdone on the gas will be :

Solution

Physics sir · 4 steps

Get P₂ from PV^γ constant, then work on the gas is the change in PV over γ - 1.

1. Final pressure. A monoatomic gas has γ = 5/3. The volume shrinks by a factor of 1200/150 = 8, and 8^(5/3) = 2⁵ = 32. So P₂ = 75 × 32 = 2400 kPa.

   W_(on gas) = (P₂V₂ - P₁V₁)/(γ - 1)

2. Put in numbers. Use 1 cm³ = 10⁻⁶ m³. Then P₁V₁ = 75 × 10³ × 1200 × 10⁻⁶ = 90 J and P₂V₂ = 2400 × 10³ × 150 × 10⁻⁶ = 360 J.
3. Work done. W = (360 - 90)/(2/3) = (270 × 3)/2 = 405 J. It is positive, as it should be, because the gas is being compressed.
4. So. The work done on the gas is 405 J, option (b).

The idea to keep. For any adiabatic process, work done on the gas is (P₂V₂ - P₁V₁)/(γ - 1). Get the missing pressure from PV^γ constant, and convert cm³ to m³ first.

Answer from the NTA final answer key. Solution by PrepWiser.