Thermodynamics Graph Questions for Class 11 Students

Question 1

Below diagram shows a process ABCA performed on an ideal gas. Find the net heat given to the system during the process.

Solution
As process ABCA is a cyclic process so net internal energy change during it should be zero.
ΔUABCA = 0

Work done during process AB
W = pΔV
ΔV = 0 because volume is constant during process AB
WAB = 0

Work done during process BC
As process BC is done at constant temperature thus it’s an Isothermal Process
WBC = nRT2 ln(V2/V1)

Work done during process CA
From the diagram, it’s clear that as volume is decreasing from C to A temperature is also decreasing. Thus during process CA, volume is directly proportional to temperature.
V ∝ T

This means that ratio V/T is a constant

Thus using ideal gas equation p = nRT/V is also constant because all of n, R, T, V are constants.

As p = nRT/V
⇒ pV = nRT

Thus
At point C ⇒ pV2 = nRT2
At point A ⇒ pV1 = nRT1

So for process CA pressure is constant.

Work done during process CA
WCA = pΔV = p(V1 – V2) = pV1 – pV2

WCA = pV1 – pV2

Replacing pV1 = nRT1 and pV2 = nRT2

WCA = nRT1 – nRT2 = nR(T1 – T2)

WCA = nR(T1 – T2)

ΔUABCA0
WAB0
WBCnRT2 ln(V2/V1)
WCAnR(T1 – T2)

Total done during process ABCA will be WAB + WBC + WCA

⇒ WABCA = WAB + WBC + WCA

WABCA = 0 + nRT2 ln(V2/V1) + nR(T1 – T2)

WABCA = nRT2 ln(V2/V1) + nR(T1 – T2)

Using First Law of Thermodynamics
ΔU = ΔQ – ΔW

For process ABCA
ΔUABCA = ΔQABCA – ΔWABCA

Replacing
ΔUABCA = 0

WABCA = nRT2 ln(V2/V1) + nR(T1 – T2)

⇒ 0 = ΔQABCA – [nRT2 ln(V2/V1) + nR(T1 – T2)]

ΔQABCA = nRT2 ln(V2/V1) + nR(T1 – T2)

Thus heat released during process ABCA can be calculated as nRT2 ln(V2/V1) + nR(T1 – T2)

Question 2

Consider the cyclic process ABCA on a sample of 2.0 mol of an ideal gas as shown in below diagram.

The temperatures of gas at A and B are 300K and 500K respectively. A total of 1200 J of heat is withdrawn from sample in the process. Find the work done by gas in part BC.
Use R = 8.3 J K-1 mol-1

Solution
Change in internal energy of an ideal gas during cyclic process is zero.
Thus
ΔUABCA = 0

Suppose that heat supplied to gas during process ABCA is ΔQABCA and work done by gas during this process is ΔWABCA

Then as per First Law of Thermodynamics
ΔUABCA = ΔQABCA – ΔWABCA

Replacing ΔUABCA = 0

0 = ΔQABCA – ΔWABCA

ΔQABCA = ΔWABCA

Thus heat supplied to gas during process ABCA is equal to work done by it.

In the question it’s given that total heat withdrawn from system is 1200 J, but with a negative sign this will become total heat supplied to system.

Thus heat supplied to gas during process ABCA = ΔQABCA = – 1200 J

As ΔQABCA = ΔWABCA and ΔQABCA = – 1200 J

ΔWABCA = – 1200 J

Question 3

Two processes A and B are shown in the below diagram, let ΔQ1 and ΔQ2 be the heat given to system in processes A and B. Then find relationship between ΔQ1 and ΔQ2.

Solution
From the First Law of Thermodynamics we have
ΔQ = ΔU + ΔW

From the diagram it’s clear that both of processes A and B are taking gas from one same state to another same state. Thus, change in internal energies for both of these processes will be same.
⇒ ΔUA = ΔUB

From the diagram, it’s also clear that area under curve for process A is bigger than area under curve for process B. Using the fact that area under a P-V diagram is work done. We can say that work done in process A is larger than that in process B.
⇒ ΔWA > ΔWB

Equations for First Law of Thermodynamics
ΔQA = ΔUA + ΔWA
ΔQB = ΔUB + ΔWB

Subtracting these two equations
ΔQA – ΔQB = ΔUA + ΔWA – (ΔUB + ΔWB)

ΔQA – ΔQB = ΔUA + ΔWA – ΔUB – ΔWB

Replacing ΔUA with ΔUB in above equation as ΔUA = ΔUB

ΔQA – ΔQB = ΔUA + ΔWA – ΔUA – ΔWB = ΔWA – ΔWB

ΔQA – ΔQB = ΔWA – ΔWB

As ΔWA > ΔWBΔWA – ΔWB > 0

Thus ΔQA – ΔQB > 0 ⇒ ΔQA > ΔQB

Question 4

When a system is taken through the process abc as shown in below diagram. 80 J of heat is absorbed by the system and 30 J of work is done by it. If the system does 10 J of work during the process adc, how much heat flows into it during the process?

Solution
So we two process taking gas from state a to c
First process is a ➜ b ➜ c and second one is a ➜ d ➜ c

ΔQa ➜ b ➜ c = 80 J
ΔWa ➜ b ➜ c = 30 J

As per First Law of Thermodynamics
ΔU = ΔQ – ΔW

Applying this on process a ➜ b ➜ c
ΔUa ➜ b ➜ c = ΔQa ➜ b ➜ c – ΔWa ➜ b ➜ c = 80 – 30 = 50 J

ΔUa ➜ b ➜ c = 50 J

From the diagram it’s clear that both process a ➜ b ➜ c and a ➜ d ➜ c are taking gas from same initial state to same final state, thus change in internal energy of gas for both of these processes should be same.

Thus
ΔUa ➜ b ➜ c = ΔUa ➜ d ➜ c = 50 J

As per question it’s given that
ΔWa ➜ d ➜ c = 10 J
ΔUa ➜ d ➜ c = 50 J

As per First Law of Thermodynamics
ΔU = ΔQ – ΔW

Applying this on process a ➜ d ➜ c
ΔUa ➜ d ➜ c = ΔQa ➜ d ➜ c – ΔWa ➜ d ➜ c

Putting in the values
50 = ΔQa ➜ d ➜ c – 10

50 – 10 = ΔQa ➜ d ➜ c = 40 J

Thus heat absorbed during process a ➜ d ➜ c is 40 J

Question 5

A gas is taken along the path AB as shown in diagram below, if 70 J of heat is extracted from the gas in the process, calculate the change in internal energy of the system.

As per question its given that
70 J of heat is extracted from gas in the process A ➜ B

Thus
ΔQ = 70 J

We know that area of a V-P curve with V axis is work done during the process.

Work Done = Area of region ABCE
= Area of triangle ACB + Area of rectangle ACDE

Area of triangle ACB = 0.5 × BC × AC
BC = 500 – 200 = 300 kPa = 300 × 103 Pa

AC = 250 – 100 = 150 cc = 150 × 10-6 m3

⇒ Area of triangle ACB = 0.5 × 300 × 103 × 150 × 10-6
(0.5 × 300 × 150) × (103 × 10-6 ) = 22500 × 10-3 = 22.5 J

Area of triangle ACB = 22.5 J

Let’s now find out Area of rectangle ACDE
= Length × Breadth = AC × CD

AC = 250 – 100 = 150 cc = 150 × 10-6 m3

CD = 200 – 0 = 200 kPa = 200 × 103 Pa

Area of rectangle ACDE = 150 × 10-6 m3 × 200 × 103 Pa
= (150 × 200) × (10-6 × 103)

= 30000 × 10-3 = 30 J

Area of rectangle ACDE = 30 J


Work Done = Area of region ABCE
= Area of triangle ACB + Area of rectangle ACDE
= 22.5 + 30 = 52.5 J

Work done during process = 52.5 J
ΔW = 52.5 J

As per First Law of Thermodynamics
ΔU = ΔQ – ΔW

ΔQ = 70 J (Given in question)

ΔW = 52.5 J (Calculated above)

ΔU = 70 – 52.5 = 17.5 J

ΔU = 17.5 J

Thus internal energy of gas increases by 70 J as it’s state is moved from A to B.




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