Question

1. A manometer is attached to a sample of gas that occupies a volume of 2.5 liters. The mercury level in the arm of the manometer attached to the gas sample is 19.5 torr higher than that of the arm open to the atmosphere. Atmospheric pressure is measured to be 0.98 atm. What is the pressure of the sample of gas? (6 pts) 2. A balloon has a volume of 2.32 liters at 24.0°C. The balloon is heated to 48.0°C. Calculate the new volume of the balloon. (4 pts) 3. The temperature of a specific amount of gas in a sealed container changes from 30.0°C to 65,0°C. If the volume remains constant, and the initial pressure was 98.89 kPa, what is the new pressure of the gas? (6 pts)

Answer 1

Palm Page No.: L. I I This h represents the ci height difference. The pressure corresponding to height difference is 19.5 tony Parm is exerting more pressure. lic Patm = gas - gas + h. (so) (no) I atm = 760 toro 74408 tooo - 19.5tory 0.98 am = 0.98X760 725.3 tony. ON - 744.8 toor Pam- h a = 2= (as Assuming pressure Apply Boglostas remaining constant Charles law. Vaa V, XT2 Vi= 2.32 L T, = 24°C =24+273 = 2978 V = 2.324 32) 297 = 9:51 SL Tz= 48°C 48+273 - 321K

Page No. : volume is we can apply gt states that gas vassies constant. Gay - Lussac Law at constant volume, pressure of dixully with temperature P = 98.89 kPa Tis 30°C = 30 +273= 303K Iz= 65°C 65+273=3385 »Р, - Р, ХТ, 98.89 x 338 303 110.31 kPa. =