1 Heat capacity of gasses

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1 1 Heat capacity of gasses 1.1 Objective Determine the molar heat capacities of air at constant volume Cv and at constant pressure Cp. 1.2 Principle and Task Heat is added to a gas in a glass vessel by an electric heater which is switched on briefly.under isochoric conditions the temperature increase results in a pressure increase, which is measured with a manometer.under isobaric conditions a temperature increase results in a volume dilation,which can be read from a gas syringe. the molar heat capacities Cv and Cp are calculated from the pressure or volume change. 1.3 Setup of the Experiment Perform the experimental set-up according to Fig.1 and 2.Insert the two nickel electrodes into two holes in the three-hole rubber stopper and fix the terminal screws to the lower ends of the electrodes. Screw two pieces of chrome nickel wire, which are each about 15 cm long, into the clamps between these two electrodes so that they are electrically connected in parallel. The wires must not touch other.insert the one-way stopcock into the third hole of the stopper and insert the thus-prepared stopper in the lower opening of the bottle. The 5 V output of the electrical 4-decade digital counter serves as the power source. The electrical circuit is illustrated in Fig Theory The first law of thermodynamics can be illustrated particularly well with an ideal gas. This law describes the relationship between the change in internal intrinsic energy U i the heat exchanged with the surroundings Q and the constant-pressure change ρdv dq = du i + ρdv (1.1) The molar heat capacity C of a substance results from the amount of absorbed heat and the temperature change per mole: 1

2 where n = numberofmoles C = 1 n dq dt (1.2) One differentiates between the molar heat capacity at constant volume C v and the molar heat capacity at constant pressure C ρ. According to equations(1)and(2)and under isochoric conditions(v constant,dv=0),the following is true: C v = 1 n du i dt and under isobaric conditions(p=constant,dp=0): (1.3) C ρ = 1 n (du i dt + ρdv dt ) (1.4) Taking the equation of state for ideal gases into consideration: ρv = nrt (1.5) it follows that the difference between C v and C ρ for ideal gases is equal to the universal gas constant R. C ρ C v = R (1.6) It is obvious from equation(3) that the molar heat capacity C v is a function of the internal intrinsic energy of the gas.the internal energy can be calculated with the aid of the kinetic gas theory from the number of degrees of freedom f: U i = 1 2 f k BN A T n (1.7) where k B = (1 38) J/K (Boltzmann Constant) N A = (6.02) mol 1 ( Avogadros number) Through substitution of R = k B N A (1.8) it follows that C v = f 2 R (1.9) and taking equation(6) into consideration: C ρ = ( f + 2 )R (1.10) 2 2

3 The number of degrees of freedom of a molecule is a function of its structure.all particles have 3 degrees of translational freedom.diatomic molecules have an additional two degrees of rotational freedom around the principal axes of inertia.triatomic molecules have three degrees of rotational freedom. 1.5 Determination of C ρ The energy Q is supplied to the gas by the electrical heater: Q = U.I. t (1.11) where: U =the voltage which is applied to the heater wires. I= the current,which flows through the heater wires. t=the period of time in which current flowed through the wires. At constant pressure the temperature increase T induces a volume increase V from the equation of state for ideal gases, it follows that: V = nr ρ T = V T (1.12) T and taking equation (2) into consideration,the following results from equations(11)and(12): C ρ = 1 U.I. t.v n V.T (1.13) The molar volume of a gas at standard pressure ρ 0 = 1013hP a and = 273.2k is: V 0 = /mol. The molar volume is: V mol = p 0v o in accordance with the following,the number of moles in volume V IS: T P (1.14) n = V V mol (1.15) C ρ can be calculated using equation(13) under consideration of(14) and (15): C ρ = p 0v o UI p t V (1.16) The pressure p used in equation (16) is calculated from the atmospheric pressure minus the pressure reduction due to the weight of the syringes plunger. The pressure reduction 3

4 is calculated from: P k = m k.g F k = kg.9.81ms m 2 = 14.8hP a P = P a P k = 975hP a 14.8hP a = 960hP a where pa =the atmospheric pressure Pk=the pressure reduction due to the weight of the plunger mk =0.1139kg=mass of the plunger g=acceleration of gravity, F k = m 2 =area of the plunger 1.6 Set up and procedure The experiment is set up as shown in Fig.2 1. Two gas syringes,are connected to the bottle from the three-way stopcock. 2. One of the gas syringes is mounted horizontally and the other is positioned vertically. 3. While making measurements,the three-way cock must be positioned in such a manner that it only connects the vertical syringe with the bottle. 4. Start the plunger rotating manually before the measurement so that it rotates throughout measurement. 5.Start the measuring procedure by activating the push-button switch. 6.The period and the corresponding volume increase( V )are read. 7. Repeat the step(6)at least 10 measurements. 8.After each measurement remove air from the system,to do this, turn the three-way cock in such a manner that both syringes and the bottle are connected with each other. 4

5 1.7 Measurements and calculations Table 1.1: The results of a typical measurements for the determination of Cp t(sec) V (cm 3 ) (a.)plot V (cm 3 ) as y-axis versus t(sec) as x-axis. (b.)from the graph find the slope of the line,where Slope = V t (C.)Calculate the value of Cp by using eq.(16),where: C ρ = p 0v o UI p 1 Slope 1.8 Determination of Cv Under isochoric conditions, the temperature increase T produces a pressure increase P. The pressure measurement results in a minute alteration of the volume which must be taken into consideration in the calculation: T = P nr V + V nr P = it follows from equations(3) and (1) that: T (P V + V P ) (1.17) P V C v = 1 n and with equations(11) and (17) one obtains: Q P V T (1.18) C = P.V n.t U.I t p V P V + V P (1.19) The indicator tube in the manometer has a radius of r=2 mm. A pressure change of p=0.147 hpa causes an alteration of 1 cm in length; the corresponding change in volume is therefore: V = a P (1.20) 5

6 where thus a = πr cm hp a = cm3 hp a (1.21) C v = P V U.I t ap. P nt.(ap + V ). P Taking equations(14) and (15) into consideration, it follows that: (1.22) 1.9 Set up and procedure The experiment is set up as shown in Fig.3 C v = p 0v o U.I t ( (ap + V ). P ap ap + V ) (1.23) 1.To determine Cv, connect the precision manometer to the bottle with a piece of tubing. 2.Start the measuring procedure by activating the push-button switch. 3.Read the time and the pressure increase immediately after cessation of the heating process. 4.Repeat the step(2) and (3)at least 10 measurements. 5.After each measurement,perform a pressure equalisation with the atmospheric pressure by opening the three-way cock. 6

7 1.10 Measurements and calculations Table 1.2: The results of a typical measurements for the determination of Cv t(sec) P (hpa) (a.)plot P (hpa) as y-axis versus t(sec) as x-axis. (b.)from the graph find the slope of the line,where Slope = P t (C.)Calculate the value of Cv by using eq.(23),where: C v = p 0v o ( U.I (ap +V ).Slope ap ) ap +V (d.)calculate R, using the relation: R = C p C v 7

Heat capacity of gases

Heat capacity of gases LEP Related topics Equation of state for ideal gases, 1st law of thermodynamics, universal gas constant, degree of freedom, mole volumes, isobars, isotherms, isochors and adiabatic