Problems on thermodynamics
Q1

A system does 164 J of work on its environment and gains 77 J of heat in the process. Find the change in the internal energy of
(a) the system
(b) the environment ?

Sol:

According to the first law of Thermodynamics change in internal energy is

ΔU = Q – W
Work done by the system W = 164 J
Heat gained by the system Q = 77 J
Change in internal energy of the system ΔU = 77 – 164
= – 87 J
This means that the internal energy of the system decreases.
Now in the case of environment
Work is done on the environment W = – 164 J
Heat is given to the system Q = – 77 J
∴ Change in internal energy ΔU = – 77 + 164
= 87 J

This means that internal energy of the environment increases.

Q2

A gas, while expanding under isobaric conditions, does 480 J of work. The pressure of the gas is 1.6 × 105 pa, and its volume is 1.5 × 10– 3 m3. What is the final volume of the gas ? –1 for all frequencies, calculate the wave lengths corresponding to the given extreme frequencies of the range ?

Sol:

In an isobaric expansion process, the pressure is constant and volume changes from V1 to V2.
Then the work done is given by

W = P(V2 – V1)
But here W = 480 J
P = 1.6 × 105 pa
V1 = 1.5 × 10– 3 m3
∴ Final volume V2 = + V1
= + 1.5 × 10 – 3
= 4.5 × 10– 3 m3
Q3

A certain gas at atmospheric pressure is compressed adiabatically so that its volume becomes half of its original volume. Find the resulting pressure. Given γ = 1.4 ?

Sol:

In an adiabatic process, the relation between pressure and volume is given by PVγ = constant

or P1V1γ = P2V2γ
Given Initial pressure P1 = 1 atm
= 1.013 × 105 N/m2
Final volume V2 = V1
γ = 1.4
Putting these values in the above equation, we get
P2 = P1
= 1.013 × 105(2)1.4
= 2.7 × 105 N/m2
Q4

Calculate the work done to isothermally compress 1g of hydrogen gas at NTP to half of its initial volume. Gas constant R = 8.31 J.mol k. Find the change in internal energy and amount of heat evolved ?

Sol:

Work done in an isothermal process is

W = 2.3026 RT log
For 1g of gas W = 2.3026 × × T log
Here, Mol.wt of the gas (H2) = 2
Gas constant R = 8.314 J/mol k
Temperature T = 273 k (NTP conditions)
Initial volume, V1 = V
Final volume, V2 =
∴ W = 2.3026 × × 273 log
= – 786.3 J

The negative sign indicates that the work is being done by the gas.
In an isothermal process, temperture remains constant.
Hence change in internal is zero.
According to first law of thermodynamics,
dQ = dU + dW
as dU = 0, the amount of heat evolved is equal to the amount of work done.
dQ = dW = 786.3 J

Q5

A cylinder contains 1 mole of oxygen at a temperature of 27° c. The cylinder is provided with a frictionless piston which maintains a constant pressure of 1 atm on the gas. The gas is heated until its temperature rise to 127° c. Calculate the work done by the gas, increase in the internal energy and the heat supplied to gas ? Cp = 7.03 cal/mol k, R = 1.99 cal/mol k

Sol:

As the pressure on the gas is maintained constant this is an adiabatic process.
Given that the temperature of the gas is increased from 27° c to 127° c

So initial temperature T1 = 27 + 273
= 300 k
Final temperature T2 = 127 + 273
= 400 k
Work done in an adiabatic process is W = PdV
but PV = RT for one mole of a gas.
∴ W = PdV
= RdT
= R(T2 – T1)
Given R = 1.99 cal/mol k
= 1.99 × 4.2 J/mol k
= 8.358 J/mol k
∴ Work done W = 8.358(400 – 300)
= 835.8 J
Change in internal energy of the gas dU = CvdT
Given Cp = 7.03 cal/mol k
= 7.03 × 4.2 J/mol k
= 29.53 J/mol k.
But Cp = Cv = R
⇒ Cv = Cp – R
= 29.53 – 8.358
= 21.168 J/mol k
∴ Increase in the internal energy of the gas
dV = CvdT
= 21.168(400 – 300)
= 2117 J
The amount of heat supplied to the gas is
dQ = CpdT for 1 mole of the gas
= 29.53(400 – 300)
= 2953 J
Q6

A typical doughnut contains 2g of protein, 17g of carbonhydrates and 7g of fat. The average food energy values of these substances are 4 kcal/g for protein and carbohydrates and 9 kcal/g for fat.
(a) During the exercise, if an average person uses an energy of 510 kcal/hr, how long would he has to exercise to work off one doughnut ?
(b) If the energy in the doughnut could somehow be converted into the kinetic energy of the body as a whole, how fast could one move after eating the doughnut ? Let the mass of the person be 60 kg.

Sol:

Given that a typical doughnut contains proteins – 2g, carbohydrates – 17g, fat – 7g
The energy values of these quantities are
Protein – 2 × 4 = 8 kcal
Carbohydrates – 17 × 4 = 68 kcal
fat – 7 × 9 = 63 kcals
Total energy a person gets on eating one dough nut is
Q = 8 + 68 + 63 = 139 kcal
(a) Given that a person uses an energy of 510 kcal/hr, that is 510 kcal per one hour
Then time taken to work off 139 kcal – ?

t =
; = 0.275 hr
; = 0.275 × 60 min
t = 16.4 min

So the person needs to exercise for 16.4 min to work off one doughnut.
(b) If energy in the doughnut is converted into kinetic energy of the person's body, then

Q = mv2
i.e., mv2 = 139 kcal
Given m = 60 kg
v2 = (583.8 × 103 J)/30
v =
v = 139 m/s
; = 139 × km/hr
; = 501 km/hr

The speed with which the person can walk after eating the doughnut is 501 km/hr

Q7

A gasoline engine takes in 1.61 × 104 J of heat and delivers 3700 J of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of 4.6 Þ 104 J/g.
(a) What is the thermal efficiency ?
(b) How much heat is discarded in each cycle ?
(c) What mass of the fuel is burned in each cycle ? what is the power output for 60 cycles/s.

Sol:

Given data is

Heat taken by the engine QH = 1.61 × 104 J
Work done per cycle W = 3700 J
Latent heat of combustion Lc = 4.6 × 104 J/g
(a) Thermal efficiency e =
=
= .2298 × 10– 4
; = 0.23 × 100
e = 23 %
(b) We know that work done = Heat input – Heat rejected
∴ Amount of heat rejected in each cycle
Qc = QH – W
; = 1.61 × 104 – 3700
; = 12400 J
Heat required to burn the gasoline is given by
QH = mLc
⇒ m =
; =
; = 0.35 g
Power output =
; = (work done) × (no of cycles)
Given number of cycles per sec
 = 60
∴ Power output = (3700 J/cycle)(60 cycles/sec)
; = 222000 W
; = 222 kW
Q8

A gasoline engine has a power output of 180 kw. Its thermal efficiency is 28 %. How much heat must be supplied to the engine per second and how much heat is discarded by the engine per second ?

Sol:
Given power output = 180 kw
∴ Work done per sec = 180 × 103 J
Thermal efficiency, e = 28 % = 0.28
∴ Heat supplied to the engine =
=
QH = 642.8 × 103 J
Heat rejected by the engine Qc = QH – W
; = 642.8 × 103 – 180 × 103
; = 462.8 × 103 J
Q9

A carnot engine whose high temperature reservoir is at 620 k takes in 550 J of heat at this temperature in each cycle and gives up 335 J to the low temperature reservoir. How much work does the engine perform during each cycle and what is the temperature of low temperature reservoir ? Find the thermal efficiency of the cycle ?

Sol:
Temperature of high temperature reservoir TH = 620 k
Heat input QH = 550 J
Heat rejected Qc, to cold reservoir Qc = 335 J
work done per cycle W = QH – Qc
= 550 – 335
= 215 J
Thermal efficiency e =
; =
; = 0.39 × 100
e = 39 %
But efficiency e = 1 –
Temperature of cold reservoir Tc = (1 – e)TH
; = (1 – 0.39)620
; = 378.2 k
Q10

A carnot heat engine has a thermal efficiency of 0.6 and temperature of its hot reservoir is 800 k. If 3000 J of heat is rejected to the cold reservoir in one cycle, what is the work output of the engine during one cycle ?

Sol:
Thermal efficiency of the carnot engine e = 0.6
Amount of heat rejected to cold reservoir Qc = 3000 J
Temperature of hot reservoir TH = 800 k
We know that efficiency e = 1 –
= (1 – e)
QH =
QH =
QH = 7500 J
Now the amount of work done can be found from the relation
W = QH – Qc
; = 7500 – 3000
W = 4500 J
Q11

Find the change in entropy when 1 kg of ice at 0° c is melted and converted to water at 0° c. Also find the change in entropy when 1 kg of water at 0° c is heated to 100° c. Heat of fusion of water Lf = 3.34 × 105 J/kg

Sol:

The melting occurs at a constant temperature 0° c is 273 k
So this is a reversible isothermal process.

Heat needed to melt 1 kg of ice is Q = mLf
= 1 × 3.34 × 105 J
∴ The increase in entropy Δs =
=
= 1.22 × 103 J/k

In the second case, it is not an isothermal process, the temperature changes fron 0° c to 100° c.
We can imagine that the temperature of water is increased reversibly in a series of infinite simal steps, in each of which the temperature is raised by an infinitesimal amount dT. We then integrate and calculate the change in entropy.
Heat required to carry out each infinite simal step is dQ = mcdT
Hence the change in entropy is = 1.31 × 103 J/k