Answer to Question #139775 in General Chemistry for Emily

Question

Divers know that the pressure exerted by the water above them increases about 100kPa with every 10.2m of depth. This means that at 10.2 m below the surface, the pressure is 201 kPa, at 20.4 m below the surface it’s 301 kPa & so forth. Given that the volume of a balloon is 3.5 L at STANDARD TEMPERATURE & PRESSURE (273 K & 1 atm), what is the volume of the balloon 51 m below the water’s surface if the temperature remains constant?

Answer 1

pV=nRT.

n=pV/RT = 101.325 x 3.5 / 8.314 x 273= 0,156247109 mol.

51 x 100/10.2 = 500 kPa.

V=nRT/p = 0,156247109 x 8.314 x 273 / 500 = 0,709275 L.