Test: Thermodynamic Processes - NEET MCQ

From the first law of thermodynamics dQ = dU + dW, so clearly for the isothermal expansion or compression of a real gas where u = f(T) from the first law dU = 0 which means that the entire heat supplied is converted into work but from the second law of thermodynamics we find that in no process can the entire heat supplied can  be converted into work hence in reality some fraction of heat supplied is always used to increase the internal energy of the system.

Extensive variable →H (enthalpy), E (Internal energy) and Mass. Since these variables depend on the amount of substance or volume or size of the system.

As work is a path function rather than a state function, we can easily say that work can often be graphically represented as the area under the PV graph. And as cyclic processes are represented as closed shapes on PV graph it is obvious that they have non zero area and thus work done is non zero.

From the first law of thermodynamics,
we know, dU = dQ - dW ; (work done BY the system is considered +ve)
For an adiabatic process, dQ = 0, and hence, dU = -dW
For an ideal gas expansion, we see that work done
BY the system is +ve (recall the sign convention for work done), i.e., dW > 0.
Therefore, dU is less than 0, and thus, the internal energy decreases.

T_Initial  = t K
Work, W = 9R
Ratio of specific heats, γ = C_p / C_v = 4/3
In an adiabatic process, we have
W = R(T_Final – T_initial) / (1-γ)
9R = R (T_Final – t) / (1 – 4/3)
T_Final – t = 9 (-1/3) = -3
T_Final  = (t-3) K

Correct Answer : C

Explanation : The adiabatic condition is given by the relation between pressure volume and temperature volume as:

(PV)γ = constant 

where, γ = Cp/Cv is ratio of the specific heats

These relations suggest that an decrease in volume is associated with increase in temperature

ATQ 

P₁(V₁)γ = (P₂V₂)γ

=> P₁(1)² = P₂(4)²

P₁/P₂ = 16

A reversible process gives the maximum work.

The symbol c stands for specific heat and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00degreeC. The specific heat c is a property of the substance; its SI unit is J/(kg⋅K) or J/(kg⋅C).

Work done in isothermal process is greater than work done in adiabativ process so P of X is greater than Y
 

Entropy is defined as dQ/T. In an isentropic process dQ/T = 0 which means dQ = 0. Diathermic is a process in which heat flow is easily possible. Isochoric process is one in which volume is constant. Isothermal process is one in which temperature stays constant.

An isobaric expansion of a gas requires heat transfer to keep the pressure constant. An isochoric process is one in which the volume is held constant, meaning that the work done by the system will be zero.

In a cyclic process, the starting position is the same as the ending position. So, the change in all state variables is zero. So the net change in internal energy is zero. ΔU = ΔQ – ΔW = 0.

In an isothermal process, PV=const. This is the same as Boyle’s law. Charle’s law is given by: V/T=const. Gay-Lussac’s law is given by: P/T=const and 2nd law of thermodynamics states that in every process total entropy of the universe must increase.

Since the process is isothermal the product PV will be constant.
P_AV_A = P_BV_B.
∴ V_B = 1*3/3 = 1m³.
Work done in an isothermal process is given by:
nRT*ln(V_B/V_A)
= P_AV_Aln(V_B/V_A)
= 3 ln(1/3)
= - 3.3 J.

In the given graph temperature remains constant with variation in volume. So the process is isothermal and PV = constant.