Handout : Ideal gas, internal energy, work and heat Ideal gas law For a gas at pressure p, volume V and absolute temperature T, ideal gas law states that pv = nrt, where n is the number of moles and R = 8.3 J mol - K - is gas constant. Let N A = 6.0 0 3 be the Avogadro number. For a gas with number of molecules N, the number of moles is n = N N A. Therefore, the ideal gas law can be expressed as pv = N R N A T or Figure : Gas in a container pv = Nk B T. The constant k B = R N A is called Boltzmann constant which is equal to.38 0 3 JK -. Kinetic theory of gas Consider a gas molecules in a rigid container as in Figure. Assumptions of ideal gas are: Gas molecules move randomly. The molecules are point-like; they have negligible volume. There are no intermolecular forces between gas molecules except during collision. All collisions are elastic. From these assumptions, it can be deduced that gas pressure is a result of collision of the molecules with the walls of the container. The volume of the gas is the volume of the container itself. Let m be the mass of each molecule, the assumptions of the ideal gas lead to a very important relation: pv = 3 Nm v, where v is the average of square of speeds of molecules: v = v + v +v 3 + v N. N
Substitution of pv in equation into equation leads to an expression v = 3k B T m. The rootmean-square speed is defined as v rms = v : v rms = 3k B T m. Note that in the above expression, m is the mass of one gas molecule. For example, H m =.67 0 7 kg, 4 He m = 4.67 0 7 kg, 6 8O m = 6.67 0 7 kg. By substituting k B = R N A, the formula for v rms can be written in terms of gas constant R: v rms = 3RT m A, where m A is the atomic mass in unit kg mol -. For example, H m A = 0.00 kg mol -, 4 He m A = 0.004 kg mol -, 6 8O m A = 0.06 kg mol -. Example Two moles of an ideal monatomic gas expand isobarically (constant pressure) from initial volume V = 0.03 m 3 to final volume V = 0.07 m 3. The pressure throughout is P =.5 0 5 Pa. Find the initial and final temperatures T and T. Example An ideal gas, density of.78 kg m -3, is contained in a container. The temperature of the gas is 73 K. The pressure of the gas is.0 0 5 Pa. Calculate v rms of this gas. Example Find the rms speed of oxygen molecules (O) at temperature 300 K.
3 Internal energy Internal energy of a system is the sum of total kinetic energy and potential energy of the system. In solid, atoms are joined by bonds that behave like springs as shown in Figure. The kinetic energy is from the vibrations of atoms and the potential energy is in the form of elastic energy in the springs. However, in ideal gas, there are no forces between the gas molecules. Hence, the potential energy is zero. This makes the internal energy of ideal gas equal to the total kinetic energy of the gas molecules. Figure : Atoms of solid joined by bonds that behave like springs Consider N gas molecules each with mass m. The total kinetic energy of the whole gas is E k = m v + v +v 3 + v N. The average translational kinetic energy E k = N E k = m v + v +v 3 + v N N = m v. From equation, v = 3pV/Nm and therefore E k = 3 pv N. From ideal gas law, pv = Nk B T, one arrives at E k = 3 k BT. The internal energy U is the total kinetic energy: U = E k = N E k = 3 Nk BT. The change in internal energy U is determined the change in temperature T = T T. If T > T, gas becomes hotter, U > 0, internal energy increases, If T < T, gas becomes cooler, U < 0, internal energy decreases, If T = T, there is no temperature change, U = 0, internal energy is unchanged.
E k k B T 4 Equipartition theorem So far only translational KE is considered. The factor 3/ in the expressions for E k and U can be thought of a factor / in each dimension of motion. Equipartition theorm states that for every degree of freedom, k B T is assigned to the average kinetic energy. Since molecules can move in 3 dimensions, there are 3 degrees of freedom. Hence, the average translational kinetic energy E k trans = 3 k BT = 3 k BT. For diatomic molecules, there are also rotational motion which contributes degrees of freedom as in Figure 3. The rotational energy about axis through the two atoms is negligible. Therefore, Figure 3: Rotational and vibrational motions of diatomic molecules E k rot = k BT = k B T. At room temperature, the diatomic molecules rotate but do not vibrate. At high temperatures, the molecules vibrate, contributing another degrees of freedom as in Figure 3. Therefore, Figure 4: Graph of E k k B T against T for diatomic ideal gas E k vib = k BT = k B T. Figure 4 shows average kinetic energy of diatomic molecules at various temperatures. It must be emphasized that the vibrational motion occurs only at high temperatures. Example Calculate the internal energy of mole of O at room temperature (300 K). Assume that O behaves like ideal gas. Example Evaluate the change in the internal energy of monatomic gas at constant pressure p =.5 0 5 Pa when the volume of the gas changes from V = 3.0 0 3 m 3 to V = 4.5 0 3 m 3.
Example Two thermally insulated vessels are connected by a narrow tube fitted with a valve that is initially closed as shown in the figure. The vessels are filled with oxygen. When the valve is opened, the gases in the two vessels mix and the temperature and pressure become uniform throughout. a) Find the pressure of the mixture. b) Find the temperature of the mixture. 5 P =.77 atm V = 5. L T = 80 K P =.35 atm V = 3.0 L T = 460 K Work Consider a gas of pressure p and volume V in Figure which undergoes expansion from volume V to V. The work done by gas is given by W by gas = p dv. The integral runs from state to. The work done by gas must not be confused with the work done on gas which is negative of the work done by gas. There are three cases to be considered for work done by gas W: Gas expands, V > V W > 0, Gas contracts, V < V W < 0, Volume is constant, V = V W = 0. Work done by gas can also be calculated from the area under graph of p against V. In Figure 5, if the gas undergoes change from state A to B, the gas expands and work done by gas is positive. If the gas undergoes change from state B to A, the gas shrinks and the work done by gas is negative. Work done is path dependent on the p-v diagram. Work done by the gas from state i to f depends on the paths as illustrated in Figure 6. Figure 4: Work done by gas causes increase in volume. Figure 5: p-v diagram Figure 6: Work done depending on the path on p-v diagram
Example From the p-v diagram, calculate work done by gas in one cycle 6 Example Ideal gas of n moles expands at constant temperature T and its volume changes from V to V. a) Show that the work done by the gas is W = nrt ln V V. b) For one mole of gas, how much work is done by gas to double the volume at temperature 300 K? Example Two moles of ideal gas expands at constant pressure p = 0 5 Pa. The volume of the gas increases from V =.0 liter to V =.5 liters. a) Calculate the change in internal energy b) How much work is done by the gas? c) How much heat is absorbed by the gas? Heat Gas can absorb or release heat. The specific heat capacity of the gas is often given per unit mole. The amount of heat required to change temperature of n moles of gas by T is Q = nc T, where c is called molar specific heat capacity. The unit of c is J mol - K -. The sign of Q is determined by the direction of heat flow. If heat enters the gas, Q is positive. If heat leaves the gas, Q is negative.