Test: Thermodynamics (October 10) - NEET MCQ

The first law of thermodynamics is just a conservation of energy.

Since AB is an isochoric process, so, no work is done. BC is isobaric process,
∴ W = PB × (VD – VA) = 240 J
ΔQ = 600 + 200 = 800 J
Using ΔQ = ΔU + ΔW
⇒ ΔU = ΔQ – ΔW = 800 – 240 = 560 J

Since  hence gas is monoatomic.

Under isothermal conditions, there is no change in internal energy.

Work done is not a thermodynamical function.

Efficiency of carnot engine (η₁) = 40% = 0.4; Initial intake temperature (T₁) = 500K and new efficiency (η₂) = 50% = 0.5.
Efficiency 

Therefore in first case,

⇒  T₂ = 0.6×500=300K

And in second case, 

We know that in adiabatic process, PV^γ = constant ....(1)
From ideal gas equation, we know that PV = nRT

....(2)
Puttingt the value from equation  (2) in equation (1),

  constant

P^{(1 – γ)} T^γ = constant

Initial temperature (T₁) = 18°C = 291 K
Let Initial volume (V₁) = V

Final volume 

According to adiabatic process, TV^γ – 1 = constant According to question,

T₁ = –23°C = 250 K,   T₂ = 100°C = 373K

ΔQ = ΔU + W
For adiabatic process, ΔQ = 0
ΔU = –W