# Thermodynamics Numericals For Class 11 Physics

## Question 1

A sample of an ideal gas is taken through the cyclic process abca (as shown in diagram below). It absorbs 50 J of heat during the part ab, no heat during bc and rejects 70 J of heat during ca.
40 J of work is done on the gas during the part bc.
1) Find internal energy of the gas at b and c if it is 1500 J at a
2) Calculate the work done by gas during the part ca.

Solution
1) Find internal energy of the gas at b and c if it is 1500 J at a

In going from point a to b, volume of gas is constant which means
ΔW = pΔV = 0
Thus no work is done by gas in going from a to b
As per question, during process of going from a to b, gas absorbs 50 J of heat
ΔQ = 50 J

As per First Law of Thermodynamics
ΔU = ΔQ – ΔW

Putting in the values
ΔU = 50 – 0 = 50 J

So in process of going from a to b, ΔU = 50 J which means internal energy at point b should be 50 J higher as compared to that on point a (1500 J, given in question itself)

Thus internal energy on point b will be 1500 + 50 = 1550 J

Let’s now find out what’s internal energy of gas at point c
For process bc
It’s given in the question that 40 J of work is done on gas
Which means
ΔW = – 40 J (work is done on gas, not by the gas)

It’s also given in the question that no heat is absorbed during process bc
ΔQ = 0

As per First Law of Thermodynamics
ΔU = ΔQ – ΔW
ΔU = 0 – (- 40) = 40 J
ΔU = 40 J

So for process b to c, change in internal energy of gas is 40 J which means internal energy of gas is 40 J higher at point c as compared to point b.

Earlier we’ve calculated internal energy at point b = 1550 J

So internal energy at point c will be 1550 + 40 = 1590 J

2) Calculate the work done by gas during the part ca.
In process ca
ΔU = Internal energy at a – Internal energy at c
Internal energy at a = 1500 J (given in question)
Internal energy at c = 1590 J (calculated above)

⇒ ΔU = 1500 – 1590 = – 90 J
ΔU = – 90 J (for process ca)

It’s given in question that for process ca, gas rejects 70 J of heat
ΔQ = – 70 J

As per First Law of Thermodynamics
ΔU = ΔQ – ΔW

Putting in the different values
– 90 = – 70 – ΔW
– 90 + 70 = – ΔW
– 20 = – ΔW
ΔW = 20 J

Thus during process ca, 20 J of work is done by gas.

## Question 2

A thermodynamic system is taken through the cycle abcda (as shown in diagram below)
1) Calculate the work done by gas during the parts ab, bc, cd and da
2) Find the total heat required by the gas during the process

Solution
1) Calculate the work done by gas during the parts ab, bc, cd and da
• Work done during ab
ΔW = pΔV
During ab process
p = 100 kPa
ΔV = Volume at b – Volume at a = 300 – 100 = 200 cm3

Converting these to SI units
p = 100 × 1000 = 105 Pa
ΔV = 200 × 10-6 m3

ΔW = pΔV = 105 Pa × 200 × 10-6 m3 = 20 J
Thus work done during process ab is 20 J

• Work done during bc
ΔW = pΔV
During bc process ΔV is zero because volume is not changing.
ΔW = pΔV = p × 0 = 0 J
Thus no work is done by gas during process bc

• Work done during cd
ΔW = pΔV
During cd process
p = 200 kPa
ΔV = Volume at d – Volume at c = 100 – 300 = – 200 cm3
ΔV = – 200 cm3

Converting these to SI units
p = 200 × 1000 = 2 × 105 Pa
ΔV = – 200 × 10-6 m3 = – 2 × 10-4 m3

ΔW = pΔV = 2 × 105 Pa × – 2 × 10-4 m3 = – 4 × 10 J = – 40 J
Thus during process cd 40 J of work is done on the gas

• Work done during da
ΔW = pΔV
During bc process ΔV is zero because volume is not changing.
ΔW = pΔV = p × 0 = 0 J
Thus no work is done by gas during process da

Summary
Work done during ab = 20 J
Work done during bc = 0
Work done during cd = – 40 J
Work done during da = 0 J

2) Find the total heat required by the gas during the process
Let’s calculate total work done during the process
ΔW = Work done during ab + Work done during bc + Work done during cd + Work done during da

ΔW = 20 – 40 = – 20 J
ΔW = – 20 J

As initial state is same as final state, gas goes from state a to b to c to d to a, thus it’s a cyclic process and total change in internal energy of gas during this process should be zero.

ΔU = 0

As per First Law of Thermodynamics
ΔU = ΔQ – ΔW
0 = ΔQ – (- 20)
0 = ΔQ + 20
ΔQ = – 20 J

Thus total heat required by gas during process will be – 20 J, which means gas will release 20 J of heat in the process from a to b to c to d to a.

## Question 3

Calculate the increase in internal energy of 1 kg of water at 100°C when it is converted into steam at some temperature and at 1 atm (100 kPa).

The density of water and steam are 1000 kg m-3 and 0.6 kg m-3 respectively.

The latent heat of vaporisation of water = 2.25 × 106 J kg-1

Solution
We know that
Density = Mass / Volume
Volume = Mass / Density

Volume of 1 kg of water = 1 kg / 1000 kg m-3 = 10-3 m-3

Volume of 1 kg of steam = 1 kg / 0.6 kg m-3 = 1.67 m-3

So change in volume as 1 kg of water converts to steam = 1.67 – 10-3 = 1.67 m-3

ΔV = 1.67 m-3
p = 100 kPa = 100 × 1000 = 105 Pa

Work done ΔW = pΔV = 105 Pa × 1.67 m-3 = 1.67 × 105 J

Thus 1.67 × 105 J of work is done by system as water converts to steam.

As per question
Latent heat of vaporisation of water = 2.25 × 106 J kg-1

Which means for converting 1 kg of water to steam 2.25 × 106 J of heat is required.
ΔQ = 2.25 × 106 J

As per First Law of Thermodynamics
ΔU = ΔQ – ΔW

Putting in
ΔQ = 2.25 × 106 J
ΔW = 1.67 × 105 J

ΔU = 2.25 × 106 – 1.67 × 105 = 105(2.25 × 10 – 1.67)
= 105(22.5 – 1.67) = 20.83 × 105

ΔU = 20.83 × 105 = 2.083 × 106 J

Thus for internal energy of 1 kg water will increase by 2.083 × 106 J when it’s converted to steam.

## Question 4

The internal energy of a monoatomic ideal gas is 1.5 nR(ΔT).

One mole of helium is kept in a cylinder of cross section 8.5 cm2.

The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of 42 J heat is given to the gas.

If the temperature rises through 2°C find the distance moved by piston.

Consider Atmospheric Pressure = 100 kPa

Solution
Change in internal energy of gas ΔU = 1.5 nR(ΔT)
n = 1 mol
R = 8.3 J K-1 mol-1
ΔT = 2 K

ΔU = 1.5 nR(ΔT) = 1.5 × 1 × 8.3 × 2 = 24.9 J
ΔU = 24.9 J

Heat given to gas = 42 J
ΔQ = 42 J

As per First Law of Thermodynamics
ΔU = ΔQ – ΔW

Putting in values
24.9 = 42 – ΔW

ΔW = 42 – 24.9 = 17.1 J

ΔW = 17.1 J

If distance moved by piston is denoted as x
Then work done

ΔW = pΔV

ΔV = Ax
A = 8.5 cm2 (Given in question itself)
Converting unit of A to SI Unit
A = 8.5 × 10-4 m2

ΔV = 8.5 × 10-4 × x m2
p = 100 kPa = 100 × 1000 = 105 Pa

ΔW = pΔV = 105 Pa × 8.5 × 10-4 × x m2
ΔW = 8.5x × 10 = 85x
ΔW = 85x

Just above we calculated ΔW = 17.1 J
Equating both of these equations

17.1 = 85x

Simplifying
x = 0.2 m = 20 cm

Thus if 42 J of heat is given to a monoatomic ideal gas (which have internal energy as 1.5 nRΔT) then a piston of cross sectional area 8.5 cm2 will move up by 20 cm.

## Question 5

A steam engine intakes 100g of steam at 100°C per minute and cools it down to 20°C.

Calculate the heat rejected by steam engine per minute.
Latent heat of vaporisation of steam = 540 cal g-1

Solution
Heat rejected during condensation of steam in one minute
= 100 g × 540 cal g-1
= 540 cal

Heat rejected during condensation of steam in one minute = 540 cal

Molar Heat Capacity of steam = 1 cal g-1 °C-1
Thus
Heat rejected during cooling per minute = 100 g × 1 cal g-1 °C-1 × (100°C – 20°C)
= 8000 cal

Heat rejected during cooling per minute = 8000 cal

Thus total heat rejected by engine per minute
= Heat rejected during condensation of steam in one minute + Heat rejected during cooling per minute
= 540 + 8000
= 8540 cal

Thus Steam Engine is rejecting 8540 cal of heat per minute.

## Question 6

A sample of 100g water is slowly heated from 27°C to 87°C.
Calculate the change in entropy of water.
Specific Heat Capacity of water = 4200 J kg-1 K-1

Solution
Heat supplied to increase temperature of sample from T to T + ΔT is
ΔQ = ms ΔT

Change in entropy
ΔS = ΔQ/T = msΔT/T

Thus if temperature goes from T1 to T2 and entropy goes from S1 to S2
Then above equation can be written as

$$S_{2} – S_{1} = \int_{T_{1}}^{T_{2}} ms \frac{dT}{T} = ms\;ln\left(\frac{T_{2}}{T_{1}}\right)$$

Putting in the values
m = 100g = 0.1 kg
s = 4200 J kg-1 K-1
T2 = 87°C = 360 K
T1 = 27°C = 300 K

S2 – S1 = 0.1 × 4200 × ln(360/300) = 420 × ln(1.2)

S2 – S1 = 420 × 0.1823 = 76.6 J K-1

Thus change in entropy of water = 76.6 J K-1

## Question 7

A heat engine operates between a cold reservoir at temperature T2 = 300 K and a hot reservoir at temperature T1.

It takes 200 J of heat from the hot reservoir and delivers 120 J of heat to cold reservoir in a cycle. What could be the minimum temperature of the hot reservoir?

Solution
Heat taken from hot reservoir Q1 = 200 J
Heat delivered to cold reservoir Q2 = 120 J

Work Done = Q1 – Q2 = 200 – 120 = 80 J

Efficiency of heat engine = W/Q1 = 80/200 = 0.40

From Carnot’s Theorem, no engine can have an efficiency greater than that of a Carnot Engine.

$$0.40 \leq 1 – \frac{T_{2}}{T_{1}} = 1 – \frac{300}{T_{1}} \\ 0.40 \leq 1 – \frac{300}{T_{1}} \\ 0.40 – 1 \leq – \frac{300}{T_{1}} \\ – 0.60 \leq – \frac{300}{T_{1}} \\ 0.60 \geq \frac{300}{T_{1}} \\ \frac{0.60}{300} \geq \frac{1}{T_{1}} \\ 0.002 \geq \frac{1}{T_{1}} \\ \frac{2}{1000} \geq \frac{1}{T_{1}} \\ \frac{1000}{2} \leq T_{1} \\ 500 \leq T_{1}$$

Thus minimum temperature of hot reservoir have to be 500 K.