Thermodynamic Processes - Free MCQ Practice Test with solutions, NEET Physics

Solution:

• The first law of thermodynamics states: dQ = dU + dW.
• In an isothermal process, the temperature remains constant.
• This means the change in internal energy (dU) is zero because internal energy is a function of temperature.
• Thus, all the heat supplied (dQ) is used to do work (dW).
• According to the second law of thermodynamics, not all heat can be converted to work, but in this ideal case, internal energy does not increase.

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.

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

To determine which gas has more pressure after compression, consider the processes:

• Gas X undergoes an isothermal process, where the temperature remains constant. According to Boyle's Law, if volume decreases, pressure increases.
• Gas Y undergoes an adiabatic process, where no heat enters or leaves the system. The pressure increase is more significant than in an isothermal process, as both temperature and pressure increase due to compression.

Therefore, Gas Y will have a higher pressure than Gas X after compression.

In an isothermal process, temperature remains constant and process equation is, PV = constant
So a graph is drawn between P and V.

Work done is given by,
δW = nRT log_e V₂/V₁
Form one mole
δW = RT log_e V₂/V₁ 

In an adiabatic process, heat cannot be exchanged between the system and surrounding so,
dQ = 0
By Ist law of thermodynamics
dQ = dU + dW
dU= –dW

This is the actual process equation
PVγ = constant

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.

The internal energy of a monatomic gas is due to linear motion only and that of the diatomic gas is due to rolling (linear rotatory) motion.

Since the process is isothermal, the product PV will be constant.
P_AV_A = P_BV_B.
Therefore, V_B = 1 × 3/3 = 1m³.
Work done in an isothermal process is given by:
W = nRT × ln(V_B/V_A) = P_AV_Aln(V_B/V_A)
= 3 × ln(1/3)
To elaborate: ln(1/3) is the natural logarithm of 1/3, which is approximately -1.0986. Therefore, 3 × -1.0986 = -3.3 J.

The cyclic process 1 is clockwise and the process 2 is anticlockwise. Therefore w₁ will be positive and w₂ will be negative area₂> area₁. Hence, the network will be negative.

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

Work done in closed path ABCA 
W_ABCA = Area of ΔABC
W_ABCA = 1/2 AB × BC
W_ABCA = 1/2 (P₁ – P₂) (V₁ – V₂)

For any state of an ideal gas. Therefore

T_c = 8T₀ 
Thus change in internal energy 
ΔU  = nC_vΔT 

= 10.5 RT₀

The work done in a thermodynamic process is equal to the area enclosed between the P-V curve and the volume axis.
Work done by the gas in the process A → B is

Work done in the process B → C is zero since volume remains constant. 
Work done on the gas in the procces C → D is 
W₂ = area DCB'A'
W₂ = DC × AD' = (5 × 10^–3) × (2 × 10⁵) = 1000J 
Work done in the process D → A is also zero.

Work done in the process B → C, W = 0 
Volume is constant and heat given to the system 
Q = 50J (given) 
Hence, by the first law of thermodynamics, the change in the internal energy is 
ΔU = (U_C - U_B) =  Q - W = 50J 
U_C = U_B + ΔU = 30 + 50 = 80J 
(ii)  For the process A → B, ΔU = U_B - U_A 
= 30Joule and W = area ABCD = DE × DA 
= 2 × 30 = 60J 
∴ Q = ΔU + W  = 30 + 60 = 90J 

For the process C → A, ΔU = UA - UC = 0 - 80
⇒ ΔU = -80 J
and W = area ACED = area ACB + area ABED
∴ W = (1/2 × AB × BC) + (DE × DA)
∴ W = (1/2 × 2 × 60) + (2 × 30) = 120 J
Since, in this process the volume decreases, the work will be negative (w = 120 Joule). That is, the work will be done on the system. Now, by the first law of thermodynamics, we will have
Q = ΔU + W = -80 - 120 = -200 J
Since it is negative, this heat is given out by the system and work done in the whole cycle
= area ABC = (1/2 × 2 × 60) = 60 J
Since the cyclic process is traced anticlockwise, the net work will be done on the system.

Let x ºF = x ºC , then using it in t ºC = 5/9 (t ºF – 32º)
x = 5/9 (x – 32)
or 9x  = 5x – 160
or x = – 40
Thus, – 40 ºC  

The work done against the force of friction 
= μR × displacement = 0.2 × 2 × 9.8 × 2 (in one second)
= (0.2 × 2 × 9.8 × 2) × 5 (in 5 second)
= 39. 2J 
Heat generated = 39.2/4.2 = 9.33 cal