Question

The working substance of a heat engine
is monoatomic ideal gas. Determine the
efficiency in percentage) of heat engine;
the cycle is shown in the figure.
P
400
2
2400
1
Po
3
Vo No
4V0​

Answer 1

Answer:

400 ans watt energy is required

Answer 2

The working substance of a heat engine

is monoatomic ideal gas. Determine the

efficiency in percentage) of heat engine. the cycle is shown in the figure.

solution : work done of pv curve, w = 1/2 × (4V₀ - V₀) × (4P₀ - P₀) = 9P₀V₀/2

work done from 1 to 2 = 1/2 × (4V₀ - V₀) × (4P₀ - P₀) + P₀ × (4V₀ - V₀)

= 15P₀V₀/2

heat will be given to the system when we go from 1 to 2.

change in internal energy from 1 to 2, ∆U = (4V₀ × 4P₀ - V₀ × P₀) = 15P₀V₀

now using first law of thermodynamics,

q = w + ∆U

⇒q = 15P₀V₀/2 + 15P₀V₀

= 22.5P₀V₀

now efficiency = Workdone /heat

= ( 15P₀V₀/2 )/(22.5 P₀V₀) × 100

= 4.5/22.5 × 100

= 20%

Therefore the efficiency of heat engine is 20%.