Boyle In a thermonuclear device,the pressure of the 0.50 L of gas within the bomb casing reaches 4.0x106atm. When the bomb casing is destroyed by the explosion, the gas is released into the atmosphere where it reaches a pressure of 1.00 atm. What is the volume of the gas after the explosion?

P1V1 = P2V2 V2 = P1V1 . V2 = (4.0 x 106 atm)(0.50 L) . V2 = 2.0 x 106 L Boyle

Gay-Lussac A sample of neon is at 89oC and 123 kPa. If the pressure changes to 145 kPa and the volume remains constant, find the new temperature, in K.

P1 = P2 T2 = T1P2 T1 T2 P1 . T2 = (362K) (145 kPa) . T2 = 427 K T1 = 89oC + 273 = 362 K P1 = 123kPa T2 = ?K P2 = 145 kPa P1 = P2 T2 = T1P2 T1 T2 P1 . T2 = (362K) (145 kPa) . (123 kPa) T2 = 427 K Gay-Lussac

Combined A gas at STP occupies 28 cm3 of space. If the pressure changes to 3.8 atm and the temperature increases to 203oC, find the new volume in mL.

V2 = (1atm)(0.028L)(n2)(476K) . P1V1 = P2V2 V2 = P1V1n2T2 n1T1 n2T2 n1T1P2 . V2 = (1atm)(0.028L)(n2)(476K) . (n1)(273K)(3.8atm) V2 = 0.0128 L = 13 mL Combined

IDEAL GAS LAW A sample of iodine gas has a volume of 25 L, a temperature of 25oC and a pressure of 953 mmHg. How many grams of iodine are in the sample?

V = 25 L T = 25oC+273 = 298K P = 953 mmHg = 1.25 atm n= ? mol PV = nRT n = PV RT V = 25 L T = 25oC+273 = 298K P = 953 mmHg = 1.25 atm n= ? mol n = (1.25 atm) (25 L) (0.08206 )(298K) n = 1.28 mol n = 1.28 mol I2 253.8 g I2 = 330g I2 1 mol I2 L∙atm K∙mol

IDEAL GAS LAW 25.0 g of methane (CH4) has a pressure of 4.44 atm at 250.oC. Find the volume occupied by the gas.

n = 25.0 g CH4 1 mol CH4 16.04 g CH4 n = 1.56 mol CH4 P = 4.44 atm T = 250.oC +273 = 523K V= ? V = (1.56 mol)(0.08206 )(523K) 4.44 atm V = 15.1 L PV = nRT V = nRT P L∙atm K∙mol