Identify the characteristics of an adiabatic process in a monoatomic gas. (A) Internal energy is constant. (B) Work done in the process is equal to the change in internal energy. (C) The product of temperature and volume is a constant. (D) The product of pressure and volume is a constant. (E) The work done to change the temperature from T₁ to T₂ is proportional to (T₂-T₁). Choose the correct answer from the options given below :

Solution

Physics sir · 2 steps

Put Q = 0 into the first law, then check each statement against PV^γ with γ = 5/3.

1. First law. In an adiabatic process no heat enters or leaves, so Q = 0. The first law Q = Δ U + W then gives W = -Δ U. The work done on the gas equals the rise in internal energy, so the two match in magnitude.

   W_(by gas) = nR(T₁ - T₂)/(γ - 1), PV^γ = constant, TV^(γ - 1) = constant

   Statement	Check	True?
   (A) U constant	U ∝ T and T changes, so U changes	No
   (B) W = Δ U in size	Q = 0 gives W = -Δ U	Yes
   (C) TV constant	The constant is TV^(2/3), not TV	No
   (D) PV constant	That is the isothermal law. Here PV^(5/3) is constant	No
   (E) W ∝ (T₂ - T₁)	W = 3/2nR(T₁ - T₂) for a monoatomic gas	Yes

2. So. Only (B) and (E) describe an adiabatic process, so the answer is option (b).

The idea to keep. Adiabatic means Q = 0, so all the work goes into or comes out of internal energy. Constant U and constant PV belong to the isothermal process.

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