(c) The absolute value of the ratio is PE gm=4.16×10 621×1067 It is justifiable to neglect the gravitational potential energy in theory. As the average kinetic energy is a statistical value for the whole system regardless the gravitational potential energy 2. A well-insulated 4.00 liter box contains a partition dividing the box into two equal volumes as shown in Figure 1. Initially, 2.00 g of molecular hydrogen gas(H2) at 300 K is confined to the left-hand side of the partition, and the other half is a vacuum (a)What is the rms speed of the particles in the gas? Insulation Thin partitio b)What is the initial pressure of the gas? (c)The partition is removed or broken suddenly, so that the gas now is contained throughout the entire box. Assume that the gas is ideal Gas Vacuum Does the temperature of the gas change? What is the change in the internal energy of the system? Fig. l (d)When the gas reaches equilibrium, what is the final pressure? eed of th 3RT 3×8315×300y2=193×10°(m) 2×10 (b)Using the equation of state for an ideal gas: PV=nRT=RT So the initial pressure of the gas is mRT2×10-3×8.315×300 =1.25×10°(P (c) The temperature is not changed. The internal energy of the system is not changed too (d) Because it is so fast that the gas reaches equilibrium. So it is an adiabatic process. For 1=300K and PV=PV, So the final pressure is 0 4×10 3. A gas initially at temperature Ti, pressure Pi and volume Vi has its pressure reduced to a final value P/via one of the following types of processes: (1)isochoric; (2) isothermal; (3)adiabatic (a)Sketch each process schematically on a P-V diagram b)In which process is the work done by the gas zero? (c)In which process is the work done by the gas greatest?(c) The absolute value of the ratio is 67 6.21 10 4.16 10 21 19 = × × = − − ave grav KE PE It is justifiable to neglect the gravitational potential energy in theory. As the average kinetic energy is a statistical value for the whole system regardless the gravitational potential energy. 2. A well-insulated 4.00 liter box contains a partition dividing the box into two equal volumes as shown in Figure 1. Initially, 2.00 g of molecular hydrogen gas (H2) at 300 K is confined to the left-hand side of the partition, and the other half is a vacuum. (a) What is the rms speed of the particles in the gas? (b) What is the initial pressure of the gas? (c) The partition is removed or broken suddenly, so that the gas now is contained throughout the entire box. Assume that the gas is ideal. Does the temperature of the gas change? What is the change in the internal energy of the system? (d) When the gas reaches equilibrium, what is the final pressure? Solution: (a) The rms speed of the particles in the gas is ) 1.93 10 (m/s) 2 10 3 8.315 300 ) ( 3 ( 1/ 2 3 3 1/ 2 = × × × × = = − M RT vrms (b) Using the equation of state for an ideal gas: RT M m PV = nRT = So the initial pressure of the gas is 1.25 10 (Pa) 2 10 2 10 2 10 8.315 300 6 3 3 3 = × × × × × × × = = − − − V RT M m P (c) The temperature is not changed. The internal energy of the system is not changed too. (d) Because it is so fast that the gas reaches equilibrium. So it is an adiabatic process. For T=300K and P1 V1 = P2V2 , So the final pressure is 6.25 10 (Pa) 4 10 1.25 10 2 10 5 3 6 3 2 1 1 2 = × × × × × = = − − V PV P 3. A gas initially at temperature Ti, pressure Pi and volume Vi has its pressure reduced to a final value Pf via one of the following types of processes: (1) isochoric; (2) isothermal; (3) adiabatic. (a) Sketch each process schematically on a P-V diagram. (b) In which process is the work done by the gas zero? (c) In which process is the work done by the gas greatest? Gas Vacuum Insulation Thin partition Fig.1