Sindh Mdcat Exclusive Course Physics Thermodynamics — Solved MCQs

Question 1

Refrigerator is an example of:

Accepted Answer

B. Second law of thermodynamics — According to the Second Law of Thermodynamics, heat will always flow spontaneously from hot to cold, and never the other way around. A refrigerator causes heat to flow from cold to hot by inputting work, which cools the space inside the refrigerator.

Question 2

If Cv = 5/2 R, Cp will be:

Accepted Answer

D. 7 / 2 R — Since Cp - Cv = Rhence , X - 5/2 R = RX = 7/2 R

Question 3

Change in entropy does not depend on:

Accepted Answer

D. Amount of the substance. — The formula for entropy is given by ΔS = Q/T, where ΔS represents the change in entropy, Q denotes the heat transfer, and T signifies the temperature of the system. Analyzing this formula, we observe that while the change in entropy is influenced by the heat transfer (both added and rejected) and the temperature, it remains unaffected by the amount of substance present. Therefore, option D is correct, as the change in entropy does not depend on the amount of the substance. Options A, B, and C are incorrect because they all involve factors that directly influence the entropy change according to the defined formula.

Question 4

If the temperature is increased from 200K to 800K, then what would be the change in pressure at constant volume?

Accepted Answer

A. Increases by factor 4 — PV=NRΔT V=constant ;R=constant; N=constant so P is directly proportional to T as the temperature increases 4 times as it increases from 200K to 800K so the pressure will also increase 4 times.Just be careful that the units (especially temperature) are absolute.

Question 5

For one mole of gas, the relation for specific heat capacity at constant pressure 'Cp' and at constant volume 'Cv' is _.

Accepted Answer

B. Cp - R = Cv — The relation between the specific heat capacities at constant pressure and constant volume is:Cp-Cv=RRearrange to get:Cp-R=Cv

Question 6

A container is divided into two equal portions. One portion contains an ideal gas at pressure P and temperature T while the other portion is a perfect vacuum. If a hole is opened between two portions:

Accepted Answer

C. There will be no change in internal energy. — This is correct. Because the gas expands into a vacuum with no heat exchange and no work done, internal energy remains unchanged. This is a key characteristic of free expansion for ideal gases.

Question 7

Thirty joules of heat flow into a system. The system in turn does 50 Joules of work. The internal energy of the system has:

Accepted Answer

D. Decreased by 20 joules — Apply 1st Law of Thermodynamics∆E = q - where,∆E = 30J - (+50) (work is done BY the system)∆E = -20Ji.e Internal energy decreased by 20 joules.

Question 8

For the adiabatic process, the first law of thermodynamics is:

Accepted Answer

D. w = -∆u — An adiabatic process is one in which no heat is gained or lost by the system. The first law of thermodynamics with Q=0 shows that all the change in internal energy is in the form of work done. This condition can be used to derive the expression for the work done during an adiabatic process. When a gas expands, it does work on the surroundings; compression of a gas to a smaller volume similarly requires that the surroundings perform work on the gas. If the gas is thermally isolated from the surroundings, then the process is said to occur adiabatically. The first law is; ΔQ=ΔU+w In an adiabatic change, q = 0, so the First Law becomes 0=ΔU+w -ΔU=w or ΔU=-wSince the temperature of the gas changes with its internal energy, it follows that adiabatic compression of a gas will cause it to warm up, while adiabatic expansion will result in cooling.

Question 9

The bicycle pump works based on:

Accepted Answer

A. 1st Law of thermodynamics — An adiabatic process is one where no heat enters or leaves a system. Here, we compress a gas adiabatically inside a bicycle pump. The work done on the gas increases its internal energy, increasing its temperature in accordance with the first law of thermodynamics.Increase in internal energy dU = dW the work done on the system.

Question 10

Absolute zero is equivalent to:

Accepted Answer

D. All of these — Absolute zero is the lowest possible temperature that can be theoretically achieved. It is equal to 0 Kelvin (K), -273.15 degrees Celsius (°C), and -459.67 degrees Fahrenheit (°F). All three scales have their zero points set at different temperatures, but they converge at absolute zero. At this temperature, all molecular motion theoretically ceases, and no heat energy can be extracted from a system.

Question 11

What happens to the pressure ‘P’ of an ideal gas, if the temperature is increased by a factor of 2 and the volume is increased by a factor of 8?

Accepted Answer

B. P decreases by a factor of 4 — PV=nRTSince n and R are constant so PV ∝ TP= T/VP’=2T/8V=T/4VComparing P and P’, we can see P’ decreases by a factor of 4.

Question 12

According to the first law of thermodynamics, ΔU = Q + W, where ΔU Is the increase in internal energy of the system, Q is the heat transferred to the system and W is the external work done by the system.Which of the following is NOT a correct expression?

Accepted Answer

D. When work is done by the system: ΔU = Q - W — There are two sign conventions:1. ∆U= Q+W: Here, work done BYthe gas/system is taken POSITIVE. 2. ∆U= Q-W: Here, work done ON the gas/system is taken NEGATIVE. The equation provided in the question follows the first convention, according to which D is incorrect.

Question 13

The quantity of heat required to raise the temperature of one mole of a substance through 1 K is called:

Accepted Answer

B. Molar specific heat — Specific heat is the amount of heat needed to raise the temperature of one kilogram of mass by 1 degree. The molar heat capacity is the heat capacity per unit amount (SI unit: mole) of a pure substance, and the specific heat capacity often called simply specific heat, is the heat capacity per unit mass of material. Option B is correct.

Question 14

How much heat is absorbed by 100 g of water when its temperature decreases from 25oC to 5oC? (Heat capacity is 4.2 J/g K)

Accepted Answer

D. -8400 — To calculate the heat absorbed by a substance, we can use the formula:q = m * C * ΔTWhere:q is the heat absorbed or released,m is the mass of the substance,C is the specific heat capacity of the substance, andΔT is the change in temperature.In this case, we are calculating the heat absorbed by 100 g of water when its temperature decreases from 25°C to 5°C, and the specific heat capacity of water is given as 4.2 J/g·K.ΔT = (Final temperature - Initial temperature) = (278 - 298°C) = -20KNote: The change in temperature is negative because the water is decreasing in temperature.Now we can plug in the values into the formula:q = 100 g * 4.2 J/g·K * (-20K)q = -8400 JThe negative sign indicates that heat is being released or lost by the water. Therefore, 100 g of water releases 84,00 J of heat when its temperature decreases from 25°C to 5°C.

Question 15

The value of the universal constant "R" is:

Accepted Answer

D. 8.314 J mole-1 K-1 — Physically, the gas constant is the constant of proportionality that relates the energy scale in physics to the temperature scale, when a mole of particles at the stated temperature is being considered. The value of R is 8.314J K−1mol−1.

Question 16

The efficiency of the Carnot's Engine working between 150°C and 50°C is:

Accepted Answer

C. 23.6% — The efficiency of a Carnot engine is calculated using the formula η% = (1 - T2/T1) × 100%, where T1 is the temperature of the hot reservoir and T2 is the temperature of the cold reservoir in Kelvin.Convert the given temperatures from Celsius to Kelvin: T1 = 150°C + 273 = 423 K and T2 = 50°C + 273 = 323 K.Substituting these values into the formula, η% = (1 - 323/423) × 100% = 23.6%.Options A, B, D, and E either apply the formula incorrectly or suggest efficiencies that are not supported by the calculated values.

Question 17

Bimetallic thermostat is an example of:

Accepted Answer

B. Thermal expansion — Bimetal Thermostats use two different kinds of metal to regulate the temperature setting. When one of the metals expands more quickly than the other, it creates a round arc, like a rainbow. As the temperature changes, the metals continue to react differently, operating the thermostat. ○ The basic working principle of a thermometer is thermal expansion. When the thermometric matter gets the heat energy then it expands and this expansion shows the increased temperature reading on the calibrated scale.○ A heat engine is a system that converts heat to mechanical energy, which can then be used to do mechanical work. ○ Isobaric Process is a thermodynamic process that takes place at constant pressure.○ Electric field is defined as the electric force per unit charge.So, options A, C, and D are wrong .

Question 18

Which process is shown in the graph between pressure and volume given below?

Accepted Answer

C. Isochoric — The graph above shows that pressure is increasing while the volume remains constant.The processes in which the volume is constant are said to be isochoric in nature.

Question 19

The law of heat exchange is used to determine:

Accepted Answer

D. Specific heat of a substance — The law of heat exchange is used to determine the specific heat because heat gain and loss by the body per unit kelvin is related to both the law of heat exchange and the specific heat of a substance.

Question 20

All the heat supplied to a system is converted into work in the _ process.

Accepted Answer

C. Isothermal — In the Isothermal process, the temperature is constant. Internal energy is a state function dependent on temperature. Hence, the internal energy change is zero and thus all the heat energy supplied is converted to work done.

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