Unit 3

Question Bank

Question 1) A gas exerts a pressure of 4 kPa on the walls of container 1.When container 1 is emptied into a 12-liter container, the pressure exerted by the gas increases to 10 kPa. Find the volume of container 1. Assume that the temperature and quantity of the gas remain constant.

Answer 1)

Given,

Initial pressure, P1 = 4kPa

Final pressure, P2 = 10kPa

Final volume, V2 = 12L

According to Boyle’s law, V1 = (P2V2)/P1

V1 = (10 kPa * 12 L)/4 kPa = 30 L

Therefore, the volume of container 1 is 30 L.

Question 2) A gas occupies a volume of 500cm3 at 0°C and 780 mm Hg. What volume (in litres) will it occupy at 80°C and 780 mm Hg?

Answer 2) Solution: Given, V1= 500 cm³

V2 =?

T1= 0°C= 0+273 = 273 K

T2= 90°C= 90+273 = 363 K

Here the pressure is constant and only the temperature is changed.

Using Charles Law,

(V1/ T1)= (V2 / T2)

500 / 273=V2 / 363

V2=500 * 363 / 273

V2= 664.83 cm3

1 cubic centimetre = 0.001 litre =1 x 10-3 litre

∴ 664.83 cubic centimetre = 664.83 x 10-3 = 0.664 litres

Question 3) A tyre containing 20 moles of air and occupying a volume of 60L loses half its volume due to a puncture. Considering that the pressure and temperature remain constant, what would be the amount of air in the deflated tyre?

Answer 3) Given,

The initial amount of air (n1) = 20 mol

The initial volume of the tyre (V1) = 60 L

The final volume of the tyre (V2) = 30 L

According to Avogadro’s law, the final amount of air in the tyre (n2) = (V2n1)/V1

= 30L * 20 moles / 60 L

The deflated tyre would contain 10 moles of air.

Question 4) Write the applications of first law of thermodynamics study flow process?

Answer 4) Application of first law to study flow process

A steady ﬂow process is one in which matter and energy ﬂow steadily in and out of an open system. In a steady ﬂow process, the properties of the ﬂow remain unchanged with time, that is, the properties are frozen in time. But, the properties need not be the same in all points of the ﬂow. It is very common for a beginner to confuse the term steady with the term equilibrium. But, they are not the same. When a system is at a steady state, the properties at any point in the system are steady in time, but may vary from one point to another point. The temperature at the inlet, for example, may diﬀer from that at the outlet. But, each temperature, whatever its value, remains constant in time in a steady ﬂow process. When a system is at an equilibrium state, the properties are steady in time and uniform in space. By properties being uniform in space, we mean that a property, such as pressure, has the same value at each and every point in the system.

A steady ﬂow is one that remains unchanged with time, and therefore a steady ﬂow has the following characteristics:

i. No property at any given location within the system boundary changes with time. That also means, during an entire steady ﬂow process, the total volume Vs of the system remains a constant, the total mass ms. Of the system remains a constant, and that the total energy content Es of the system remains a constant.

Ii. Since the system remains unchanged with time during a steady ﬂow process, the system boundary also remains the same

Iii. No property at an inlet or at an exit to the open system changes with time. That means that during a steady ﬂow process, the mass ﬂow rate, the energy ﬂow rate, pressure, temperature, speciﬁc (or molar) volume, speciﬁc (or molar) internal energy, speciﬁc (or molar)enthalpy, and the velocity of ﬂow at an inlet or at an exit remain constant.

Iv. Rates at which heat and work are transferred across the boundary of the system remain unchanged.

The application of the first law of thermodynamics to steady-flow processes is defined by two fundamental principles:

(a) Steady mass flow rate m = pAu (mass continuity) where p = fluid density

A = duct area

u = fluid velocity

And m = steady mass flow rate (dm/dt)

(This is usually recast as mv = uA since v = 1/p.)

(b) Conservation of energy (the steady-flow energy equation or SFEE)

Q 1-2 –W 1-2 =m [(h 2 -h 1) + ½ (u 2 2 –u 1 2) + g(Z 2 -Z l )]

Where Z = height above datum.

Note that very often gz is negligible compared with other terms and the reduced SFEE is written as

Q 2 -W 2 =m [(h2 –h1) + ½ (u 2 2 –u 1 2 ) ]

In per unit mass flow rate.

Q/m=q, W/m= w

q 2 - w 2 = (h 2 – h 1) + ½ (u 2 2 –u 1 2 )

An even further reduction by negligible kinetic energy terms here

q2 - w2 = (h2 – h1)

Many processes in reality approximate closely to steady flow, example steady

Conditions in a steam power plant after the start up transient is over and

Steady conditions exist at all points in the system.

Here two essential characteristics of steady flow:

(a) m is constant,

(b) The properties at any station are invariable with time .

The one obvious application of the SFEE where gz is prominent is in a hydroelectric plant where a change in z is essential for power production.

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Answer 5) the Difference Between Isothermal and Adiabatic Process

Isothermal process | Adiabatic process |

An isothermal process is defined as one of the thermodynamic processes which occur at a constant temperature | An adiabatic process is defined as one of the thermodynamic processes which occur without any heat transfer between the system and the surrounding |

Work done is due to the change in the net heat content in the system | Work done is due to the change in its internal energy |

The temperature cannot be varied | The temperature can be varied |

There is a transfer of heat | There is no transfer of heat |

Question 6) Explain Constant temperature

Answer 6)

Suppose a gas enclosed in the piston cylinder assembly is allowed to expand from P1 to P2 while the temperature is held constant. Then application of the first law gives:

DU = dQ – dW = dQ –PdV

It is not possible to calculate work and heat interactions unless the relationships between the thermodynamic properties of the gas are known. Suppose the gas under consideration is an ideal gas (which follows the relation Pv = RT and u =u (T) only) then for an isothermal process,

DU = 0

DQ = PdV = R.T.dv /v or Q =W = RTln(v2/v1)

Question 7) Explain Constant volume (Isochoric process)?

Answer 7)

Suppose a gas enclosed in a rigid vessel is interacting with the surroundings and absorbs energy Q as heat. Since the vessel is rigid, the work done W due to expansion or compression is zero. Applying the first law, we get

DU = dQ or Q = U2 –U1

That is, heat interaction is equal to the change in internal energy of the gas. If the system contains a mass m equal of an ideal gas, then

Q = ΔU = mCv (T2 –T1)

The path followed by the gas is shown on a P-V diagram. Now consider the fluid contained in a rigid vessel as shown. The vessel is rigid and insulated. Shaft work is done on the system by a paddle wheel as shown in Fig. a. In Fig. b electric work is done on the system. Since the vessel is rigid, the PdV work is zero. Moreover, the vessel is insulated and hence dQ = 0. Application of the first law of thermodynamics gives

DU = dQ – dW = dQ – (dWpdv + dWs)

Or dU = -dWs or – Ws = ΔU = U2 –U1

Where dWpdv is the compression /expansion work and dWs is the shaft work. That is increase in internal energy of a system at a constant volume, which is enclosed by an adiabatic wall, is equal to the shaft work done on the system.

Question 8) Explain adiabatic process formula?

Answer 8) Adiabatic process derivation

The adiabatic process can be derived from the first law of thermodynamics relating to the change in internal energy dU to the work dW done by the system and the heat dQ added to it.

DU =dQ-dW

DQ =0 by definition

Therefore, 0=dQ=dU +dW

The word done dW for the change in volume V by dV is given as PdV.

The first term is specific heat which is defined as the heat added per unit temperature change per mole of a substance. The heat that is added increases the internal energy U such that it justifies the definition of specific heat at constant volume is given as:

Cv =dUdT1n

Where,

n: number of moles

Therefore, 0= nCvdT+PdV …..(eq.1)

From the ideal gas law, we have

NRT =PV …..(eq.2)

Therefore, nRdT = PdV+VdP…. (eq.3)

By combining the equation 1. And equation 2, we get

−PdV= nCvdT = Cv / R (PdV+VdP)

0 = (1+Cv / R) PdV+Cv / RVdP

0= R+ Cv / Cv (dV / V) +dP /P

When the heat is added at constant pressure Cp, we have

Cp=Cv +R

0=γ (dV / V)+dP / P

Where the specific heat ɣ is given as:

γ ≡ Cp /Cv

From calculus, we have,

d (lnx)=dx / x

0=γd (lnV) + d (lnP)

0=d (γlnV+lnP) =d (lnPVγ)

PVγ=constant

Question 9) Consider 1 kg1 kg of air at 32 C that is expanded by a reversible polytropic process with n=1.25 until the pressure is halved. Determine the heat transfer.

Specific heat constant volume for air is 0.1786 kJ/kg.K

Answer 9)

Cn=Cv (n−k / n−1)

Where k is the heat capacity ratio.

Air is considered a diatomic ideal gas, so k = 1.4.

Cv =0.718kJ/kg∗K.

We know that

T2=T1(p2/p1)1−1/n

=305K∗ (0.5)0.2=265.52

Cn =0.718(−0.15/.25)

=−0.4308kJkg∗K

Q=m x Cn (T2−T1) =1kg∗(−0.4308)∗(265.5K−305K)=17.02kJ

Question 10) The initial volume of the gas is 5L and final volume is 3L Calculate the final pressure of the gas, given that the initial temperature is 273 K, the final temperature is 200 K, and initial pressure is 25 kPa.

Solution

According to the given parameters,

Pi= 25 kPa

Vi = 5L

Vf = 3L

Ti = 273K

Tf = 200K

According to combined gas law,

PiVi/Ti = PfVf/Tf

Substituting in the formula, we get

25 x 5 / 273 = Pf x 3 / 200

Therefore, Pf = 30.525 kPa