Boyleʼs Law: It states that volume of a given mass of a perfect gas varies inversely as the absolute pressure when temperature is constant.

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1 Properties of Pure Substance: A pure substance is one that has a homogeneous and invariable chemical composition even though there is a change of phase. In other words, it is a system which is (a) homogeneous in composition, (b) homogeneous in chemical aggregation. Examples : Liquid, water, mixture of liquid water and steam, mixture of ice and water. The mixture of liquid air and gaseous air is not a pure substance. (a) (b) Real And Perfect Gases: An ideal gas is defined as a gas having no forces of intermolecular attraction. The gases which follow the gas laws at all ranges of pressures and temperatures are considered as ideal gases. However, real gases follow these laws at low pressures or high temperatures or both. This is because the forces of attraction between molecules tend to be very small at reduced pressures and elevated temperatures. The Equation of State For A Perfect gas: Boyleʼs Law: It states that volume of a given mass of a perfect gas varies inversely as the absolute pressure when temperature is constant. If p is the absolute pressure of the gas and V is the volume occupied by the gas, then P 1 V 1 = P 2 V 2 = PV= Constant, the temperature is constant

2 Charle s law: It states that if any gas is heated at constant pressure, its volume changes directly as its absolute temperature. If a gas changes its volume from V 1 to V 2 and absolute temperature from T1 to T2 without any change of pressure, then V 1 = V 2 = V T Constant, pressure is constant Any equation that relates the pressure, temperature, and specific volume of a substance is called an equation of state. Property relations that involve other properties of a substance at equilibrium states are also referred to as equations of state. There are several equations of state, some simple and others very complex. The simplest and best-known equation of state for substances in the gas phase is the ideal-gas equation of state. This equation predicts the P-v-T behavior of a gas quite accurately within some properly selected region. To derive the equation of state for a perfect gas let us consider a unit mass of a perfect gas to change its state in the following two successive processes Formulation of equation of state

3 1. The Process (1 2), charleʼs law V 1 = V 2 We may write ( = T 3 ) V 1 = V 2 T (1) 2. The process (2 3), Boyleʼs low P 2 V 2 = P 3 V 3 We may unite (P 1 = P 2 ) P 1 V 2 = P 3 V 3 V 2 = P 3V 3 P (2) Substituting the value of (V 2 ) from equation (2) in equation (1),we get V 1 = P 3V 3 P 1 T 3 V 1 P 1 = P 3V 3 T 3 = constant vp T = constant, Where (v) the specific volume at (1 kg) vp T = R v P = RT (ideal-gas equation of state) The constant,r, is called gas constant the unites of (R) are kj has a different gas constant. The gas constant is determined as: kg.k.each perfect gas

4 R = R u M w Where : R u = the universal gas constant ( kj / kmole. K) M w = the molar mass or molecular weight (kg / kmole) The ideal gas equation of state can be written: PV = m R T For a mass (m), (ideal-gas equation of state) Where: P = The absolute pressure (Pa, k Pa) T= The absolute temperature(k) V= Volume (m 3 ) m = mass(kg) Enthalpy: (H) One of the fundamental quantities which occur invariably in thermodynamics is the sum of internal energy (u) and pressure volume product (pv). Enthalpy is very useful thermodynamic property for the analysis of engineering systems. Mathematically, it is given as, H = U+PV kj On unit mass basis, the specific enthalpy could be given as, h = u + pv kj kg Relationship between the Specific heats: Jouleʼs law: States that the internal energy of a perfect gas is a function of the absolute temperature only. u = u(t)

5 for perfect gas h = u + pv pv =RT Since R is constant and u = u(t), therefore the enthalpy is a function of temperature only too. h = h(t) since u and h depend only on temperature for an ideal gas, the specific heats C V and C p also depend,at most on temperature only. also, C V = ( u T ) v u = C V T by integration u = C V T (1) u = u 2 u 1 = C V ( ) C P = ( h T ) P h = C P T by integration h = C P T (2) h = h 2 h 1 = C P ( ) A special relationship between C P and C V for ideal gases can be obtained by differentiating the relation h = u + RT,Which yields. dh = du + R dt C P dt = C V dt + R dt C P = C V + R

6 R = C P C V The ratio of specific heat at constant pressure to specific heat at constant volume is called specific heat ratio or adiabatic index (k), k > 1 k = C P C V Some useful relations can be de derived as: C V = R K 1 C P = K R K 1 Ideal-gas specific heats of various common gases

7 The First Law of Thermodynamics: the law of conservation of energy, the net work done by the system is equal to the net heat supplied to the system. The First Law of Thermodynamics can, therefore, be stated as follows: "When a system undergoes a thermodynamic cycle then the net heat supplied to the system from the surroundings is equal to net work done by the system on its surroundings" dq = dw shows the cycle Where represents the sum for a complete cycle. Q in + ( Q out ) = W o + ( W in ) A closed system does not involve any mass flow across its boundaries: the energy balance for a cycle can be expressed in terms of heat and work interactions as: W net,out = Q net,in or W net,in = Q net,out for a cycle

8 The first law of thermodynamics (energy balance relation) in that case for a closed system becomes: Q net,in W net,out = E system (The General Energy Equation) E system = U + KE + PE For closed and stationary systems, no change in kinetic and potential energies, thus: KE = 0, PE = 0 Q net,in W net,out = U (Stationary systems) Examples: 1. The pressure gage on a 2.5m 3 oxygen tank reads 500 kpa. Determine the amount of oxygen in the tank if the temperature is 28 C and the atmospheric pressure is 97 kpa. The gas constant of oxygen is, R= kj / kg. K. Solution : P abs = P g + P atm P abs = = 597 k Pa T(k) = = 301 K PV = m R T m = PV RT = = kg A 20 m 3 tank contains nitrogen at 23 C and 600 k Pa. Some nitrogen is allowed to escape until the pressure in the tank drops to 400 k Pa. If the temperature at this point is 20 C, determine the amount of nitrogen that has escaped. The gas constant of nitrogen is, R = kj /kg. K. Assume atmospheric pressure is Kpa. = = 296 K P 1 = = Kpa = = 293 K

9 P 2 = = Kpa m 1 = P 1V = = kg R m 2 = P 2V = = kg R m = m 1 m 2 = = kg (escaped) 3. A tank contains argon at 600 C and 200 kpa gage. The argon is cooled in a process by heat transfer to the surroundings such that the argon reaches a final equilibrium state at 300 C. Determine the final gage pressure of the argon. Assume atmospheric pressure is 100 k Pa. Solution: T(k) =T( ) = = 873K = = 573K P abs = P g + P atm P 1 = = 300 kpa P 1 V 1 = P 2V 2 P 1 = P 2 At Constant volume P 2 = (573) 300 = kpa 873 P g,2 = P 2 P atm = = kpa

10 4. A piston cylinder device contains 0.8 kg of an ideal gas. Now, the gas is cooled at constant pressure until its temperature decreases by 10 C. If 16.6 kj of compression work is done during this process, determine the gas constant and the molar mass of the gas. Also, determine the constant volume and constant-pressure specific heats of the gas if its specific heat ratio is Solution: The work done by the fluid W = P (V 2 V 1 ) (1) In a constant pressure process, (P 1 = P 2 = P 3 ) PV = m RT V (1 or 2) = m R T P (2) Sub eq (2) in eq (1) we get W = m R ( ) W = R m T 16.6 = R 0.8 ( 10) R = kj kg.k ( the gas constant) M = R u R = = kg / kmol ( the molar mass) C V = R K 1 = = kj kg. C P = C V + R = = kj kg.

11 5. A perfect gas at pressure of 750 k Pa and 600 K is expanded to 2 bar pressure. Determine final temperature of gas if initial and final volume of gas are 0.2 m 3 and 0.5 m 3 respectively. Solution: Using perfect gas equation P 1 V 1 = P 1V = = = 400 K 6. A cylindrical vessel of 1 m diameter and 4 m length has hydrogen gas at pressure of 100 kpa and 27ºC. Determine the amount of heat to be supplied so as to increase gas pressure to 125 kpa. For hydrogen take C P = kj/kg. K, C V = kj/kg K.. Solution: Assuming hydrogen to be perfect gas P 1 V 1 = P 2V 2, V 1 = V 2 = P 2 = = 375 k P R = C P C V = = kj / kg. K P 1 V 1 = m R T m = (0.5) = 0.253kg Q = m C V ( ) Q = ( ) = kj (Heat added) Noting: all the heat supplied ( +Q) in a constant volume process goes to increase the internal energy. If ( Q) it means the heat rejected from the system equals the decrease in internal energy.

12 7. A gas at 65 kpa, 200 C is heated in a closed, rigid vessel till it reaches to 400 C. Determine the amount of heat required for 0.5 kg of this gas if internal energy at 200 C and 400 C are 26.6 kj/kg and 37.8 kj/kg respectively. Solution: First law of thermodynamics Q = U + PE + KE +W For closed and stationary systems ( PE = 0, KE = 0) The vessel is rigid (W = 0) Q = m (u 2 u 1 ) = 0.5 ( ) = 5.6 kj 8. A closed system of mass 5 kg undergoes a process in which there is work of magnitude 9 kj to the system from the surroundings. The elevation of the system increases by 700 m during the process. The specific internal energy of the system decreases by 6 kj/kg and there is no change in kinetic energy of the system. The acceleration of gravity is constant at g = 9.6 m /s 2. Determine the heat transfer. Solution: Applying the energy balance (first law of thermodynamics) Q W = U + PE + KE No change in kinetic energy ( KE = 0) Q = m u + m g Z +W Q = 5 ( 6) ( 9) Q = kj

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