First Law of Thermodynamics MCQ Quiz - Objective Question with Answer for First Law of Thermodynamics - Download Free PDF

​Calculation:

Given: V·T³ = constant. ⇒ V ∝ T^-3

Also given: P ∝ T⁴

From ideal gas law: PV = nRT ⇒ P ∝ T / V

Substitute V from above: P ∝ T / T^-3 = T⁴

So the relation holds.

In an adiabatic process, PV^γ = constant

⇒ P = constant × V^-γ

But we also have P ∝ T⁴ and V ∝ T^-3

Substitute into P ∝ V^-γ:

⇒ T⁴ ∝ (T^-3)^-γ = T^3γ

So, 4 = 3γ ⇒ γ = 4 / 3 ≈ tan(53°)

Final Answer: γ = tan(53°) (Option 3)

Explanation:

First Law of Thermodynamics:

• The First Law of Thermodynamics is a fundamental principle of energy conservation. For a closed system undergoing a process (not necessarily a cycle), it is expressed mathematically as:

ΔU = Q − W

Here:

• ΔU: Change in the internal energy of the system (measured in joules, J).
• Q: Heat added to the system (measured in joules, J).
• W: Work done by the system (measured in joules, J).

This equation implies that any energy added to a system in the form of heat (Q) is used to either increase the system’s internal energy (ΔU) or perform work on the surroundings (W). In other words, the energy supplied to a system is conserved and cannot be created or destroyed, only transformed from one form to another.

When a closed system undergoes a process:

1. Heat Transfer (Q): Heat may be added to or removed from the system. If heat is added, Q is positive, and if heat is removed, Q is negative. This term represents the thermal energy exchanged between the system and its surroundings.
2. Work Done (W): Work is done by the system on its surroundings or vice versa. If the system does work on its surroundings (e.g., expansion of gas), W is positive. If work is done on the system by the surroundings (e.g., compression of gas), W is negative.
3. Change in Internal Energy (ΔU): Internal energy is the total energy contained within the system, including the kinetic and potential energies of molecules. The change in internal energy reflects the net effect of heat transfer and work done.

For example, in a gas contained within a piston-cylinder arrangement:

• If heat is supplied to the gas, the gas molecules gain energy, causing either an increase in internal energy (ΔU) or expansion (work W).
• During compression, work is done on the gas, increasing its internal energy or causing heat to be rejected to the surroundings.

The equation ΔU = Q − W is valid for all closed systems undergoing a thermodynamic process, whether it involves heating, cooling, compression, or expansion.

Concept:

Apply the first law of thermodynamics:

\( \Delta U = Q - W \)

Given:

• Initial volume: \( V_1 = 0.3~m^3 \)
• Final volume: \( V_2 = 0.15~m^3 \)
• Pressure: \( P = 0.105~MPa = 105~kPa \)
• Heat rejected: \( Q = -37.6~kJ \)

Step 1: Work done

\( W = P (V_2 - V_1) = 105 \times (-0.15) = -15.75~kJ \)

Step 2: Change in internal energy

\( \Delta U = -37.6 - (-15.75) = -21.85~kJ \)

Explanation:

Given Data:

• Mass of the object, m = 5 kg
• Vertical distance fallen, h = 20 m
• Final velocity of the object, v = 10 m/s
• Initial velocity of the object, u = 0 m/s (since it falls from rest)
• Gravitational acceleration, g = 9.8 m/s²

Step 1: Calculate the potential energy lost by the object:

The potential energy (P.E.) lost by the object is given by:

P.E. = m × g × h

Substitute the given values:

P.E. = 5 × 9.8 × 20 = 980 J

Thus, the object loses 980 J of potential energy as it falls.

Step 2: Calculate the kinetic energy gained by the object:

The kinetic energy (K.E.) gained by the object is given by:

K.E. = 0.5 × m × v²

Substitute the given values:

K.E. = 0.5 × 5 × (10)² = 0.5 × 5 × 100 = 250 J

Thus, the object gains 250 J of kinetic energy during the fall.

Step 3: Calculate the work done by air resistance:

The work done by air resistance is equal to the difference between the potential energy lost and the kinetic energy gained:

Work done by air resistance = Potential energy lost - Kinetic energy gained

Work done by air resistance = 980 - 250 = 730 J

Since air resistance opposes the motion of the object, the work done by air resistance is negative:

Work done by air resistance = -730 J

Calculation:

For adiabatic process,

⇒ dQ = 0

⇒ Molar heat capacity = 0

⇒ From first law, dQ = dU + dW

⇒ 0 = dU + dW ⇒ dU = -dW

Also,

⇒ dU = (f / 2) × nR × dT

∴ Only option (3) is correct.

Explanation: 

The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. 

• The law of conservation of energy states that the total energy of an isolated system is constant. 
• Energy can neither be created nor be destroyed but can be transformed from one form to another.
   

The first law is often formulated by stating that the change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work done by the system on its surroundings

δQ = ΔU + δW

ΔU = δQ - δW

According to the First Law of thermodynamics, “For a closed system undergoing a cycle, net heat transfer is equal to network transfer.”

ΣQ = ΣW.

∴ Option (3) is the Correct Answer.

Concept:

\(W = \mathop \smallint \limits_{{v_1}}^{{v_2}} pdV = P\left( {{V_2} - {V_1}} \right)\)

Work Done = Area under P – V diagram

Calculation:

Given:

Work done = Area of Parallelogram ABCDA

Area of parallelogram = 1/2 × (Sum of parallel sides) × height

Work done = Area of Parallelogram ABCDA = 1/2 × (AB + CD) × BC

⇒ 1/2 × (V + 2V) × P = 1.5PV

Concept:

The change in internal energy is given by, \(Δ U = m{c_v}Δ T\)

Calculation:

Given:

V₂ = 2V; V₁ = V, and P₁ = P₂ = P

\(Δ U = m\frac{R}{{\gamma - 1}}\left( {{T_2} - {T_1}} \right)\)

where \({C_v} = \frac{R}{{\gamma - 1}}\) and ΔT = T₂ – T₁

\(Δ U = \frac{1}{{\gamma - 1}}\left( {mR{T_2} - mR{T_1}} \right)\)

As we know from the ideal gas equation PV = mRT.

\(Δ U = \frac{1}{{\gamma - 1}}\left( {{P_2}{V_2} - {P_1}{V_1}} \right)\)

\(Δ U = \frac{P}{{\gamma - 1}}\left( {{}{2V} - {}{V}} \right)\)

∴ we get, \(Δ U = \frac{{PV}}{{\gamma - 1}}\).

Mistake Points

In the Isobaric process, pressure is constant throughout the process.

ΔW = P₂V₂ - P₁V₁ = mR(T₂ -T₁)

ΔQ = mC_p(T2 -T1)

ΔU = mC_v(T2 -T1)

Hence for isobaric process also, the change in internal energy is given by ΔU = mCv(T2 -T1)

Concept:

The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. The law of conservation of energy states that the total energy of an isolated system is constant. Energy can be transformed from one form to another but cannot be created or destroyed.

The first law is often formulated by stating that the change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work done by the system on its surroundings

δQ = ΔU + δW

ΔU = δQ - δW

According to the First Law of thermodynamics, “For a closed system undergoing a cycle, net heat transfer is equal to network transfer.”

ΣQ = ΣW

Net work done = Net heat in the cycle

Calculation:

Given:

Q1 = 20 kJ, Q2 = - 28 kJ, Q3 = - 2 kJ, Q4 = 40 kJ

Net work done in the cycle = Net heat in the cycle

Wnet = Q1 + Q2 + Q3 +  Q4 

= 20 - 28 - 2 + 40

= 30 kJ

The total work for this cycle process is 30 kJ.

Explanation:

The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other.

This law is the basis for the temperature measurement.

• By replacing the third body with a thermometer, the Zeroth law can be restated as two bodies are in thermal equilibrium if both have the same temperature reading even if they are not in contact.
• The thermometer is based on the principle of finding the temperature by measuring the thermometric property.

Concept:

According to the first law of thermodynamics

δQ = dU + δW

Isobaric work done

δW = PdV = P (Vfinal - Vinitial) 

Calculation:

Initial condition ⇒ P₁ = 1 MPa, V₁ = y m³

Final condition ⇒ P₂ = 1 MPa, V₂ = 0.2 m³

Heat Transfer = -40 kJ (from the gas)

Change in Internal energy (u₂ – u₁) = -20 kJ (drop)

According to first law of thermodynamics

δQ = dU + δW

-40 = -20 + δW

⇒ δW = -20 kJ          ---(I)     

Since, the process is isobaric (as pressure remains same)

So, isobaric work done δW = PdV = P (V_final - V_initial) 

δW = P (V_final - V_initial) = -20 kJ

1000 kPa × (0.2 – y) m³ = -20 kJ

\(0.2 - y = \frac{{ - 20}}{{1000}} = - 0.02 \Rightarrow y = 0.22\)

Concept:

Apply the first law of thermodynamics

dQ = dU + dW

Calculation:

Given:

V₁ = 1 m³, V₂ = 2 m³, dQ = 2000 kJ, P = 1000 V kPa ⇒ 1000 × 10³ V Pa.

Heat is added to the system, so it is positive

We know work done

\(\begin{array}{l} dW = \int\limits_{\mathop V\nolimits_1 }^{\mathop V\nolimits_2 } {PdV} \\ dW= \int\limits_1^2 {1000 × \mathop {10}\nolimits^3 VdV} \\ dW= 1000 × \mathop {10}\nolimits^3 \left[ {\mathop {\left( {\frac{{\mathop V\nolimits^2 }}{2}} \right)}\nolimits_1^2 } \right]\\dW = 1000 × \mathop {10}\nolimits^3 × \left[ {\frac{{4 - 1}}{2}} \right]\\dW = 1500~kJ \end{array}\)

We know that

dU = dQ - dW

dU = 2000 - 1500

dU = 500 kJ.

Concept:

Pump:

• A pump is a mechanical device used to force a fluid (a liquid or a gas) to move forward inside a pipeline or hose.
• They are also used to produce pressure by the creation of a suction (partial vacuum), which causes the fluid to rise to a higher altitude.

The efficiency of the pump, \(\eta =\frac{Useful ~power}{Power~rating}\) 

Calculation:

Given:

Power rating = 400 W, Mass of water lifted = 500 kg, H = 30 m, time = 10 minutes = 60 × 10 = 600 seconds, g = 10 m/s²

Force, F = m × g = 500 × 10 = 5000 N

Work = Force × displacement (H) = 5000 × 30 = 150 kJ

\(Power=\frac{Work}{Time}\)

\(Power=\frac{150}{600}=0.25 ~kW=250~W\)

⇒ Useful work = 250 W

\(\eta=\frac{Useful ~power}{Power~rating}\)

\(\eta=\frac{250}{400}=0.625\) 

⇒ η = 62.5%

Concept:

When Q joule heat is added to a body whose mass is m, the temperature rises from T₁ to T₂.

It is given by Q = mc(T₂ - T₁) where c = specific heat of the body.

Calculation:

Given:

m = 2 kg, Q = 500 kJ, T₂ = 200 °C, T₁ = 100 °C   

∵ Q = mc(T₂ – T₁)

⇒ 500 = 2 × c × (200 – 100)

⇒ c = 2.5 kJ/kg°K

Important Points

When difference of temperature is needed do not convert °C into °K.

Concept:

If the balloon containing the ideal gas is initially kept in an evacuated and insulated room. Then if the balloon ruptures and the gas fills up the entire room, the process is known as free or unrestrained expansion.

Now if apply the first law of thermodynamics between the initial and final states.

\(Q=(u_2-u_1)+W\)

In this process, no work is done on or by the fluid, since the boundary of the system does not move. No heat flows to or from the fluid since the system is well insulated.

\(u_2-u_1=0\Rightarrow u_2=u_1\)

Enthalpy is given as 

h = u + Pv

For ideal gases, as we know, internal energy and enthalpy are a function of temperature only, so if internal energy U remains constant, temperature T also remains constant which means enthalpy also remains constant.

So, during the free expansion of an ideal gas, both internal energy and enthalpy remain constant.