Question

11 Review Constants Periodic Table This figure (Figure 1)shows a container that is sealed at the top by a movable piston Inside the container is an ideal gas at 1.00 atm, 20.0°C, and 1.00 Ls. This information will apply to all parts of this problem A B, and Part A What will the pressure inside the container become if the piston is moved to the 2.20 L mark while the temperature of the gas is kept constant? Express your answer with the appropriate units. View Available Hi Figure 1 of 1 Submit -2L piston - container Part B FIL The gas sample has now returned to its original state of 1.00 atm, 20.0°C and 1.00 L. What will the pressure become if the temperature of the gas is raised to 200.0°C and the piston is not allowed to move? Express your answer with the appropriate units. View Available Hint(s) ideal gas sample C) ?

Answer 1

Part A : P = 0.45 atm

eliven for an ideal gas : P = 1:00 atm (Pressure) T-20.0'c ( Temperature) V - 1.00 L (volumes Past A; If pistan is Pressure = ? hi If piston is moved to 2.20L. le now volume becomes 2.20L. Initial volume (V) = 1.00L final volume (V2) = 2.20 Initial pressure (Pi) = 1:00 atm final pressure (P2) = ? According to Boyle's law; Piv, = PzV2 (Putting all values) (1-00 aton) (1.204) = 12 (2.201) P2 = 1 = 0.45 atm 2.20 T IP = 0.45 atm Part B ; Now conditions are; P = 4:00 atm, T= 20.0°C, V=1.09 L If temp is raised to 200°C; pressure = ? According to Gay-Lussae's law : -P - Acc. to given conditions ; P = latin, T, = 20°C = 20+ 273= 293k T2 = 200°C - 473k] P2 = ? 1 atm - P2 293 k 473k 12 = 473 - 1161 atm 1P = 1.61 at 1 293-

Part B : P = 1.61 atm