A sample occupies 68.0 mL at 40.0 C and 95.0 kPa. what pressure will it have if cooled to -20.0 C while the volume remains constant?

76.703 kPa is the pressure if a sample occupies 68.0 mL at 40.0 C and 95.0 kPa and cooled to -20 degrees while volume remains constant.

Explanation:

Volume of the gas V1 = 68 ml

Temperature of the gas T1=40 degrees or 313.15 K

Pressure of the gas P1 = 95 Kpa OR 0.937 atm

Final temperature of the gas  T2 = -20 Degrees OR 253.15 K

volume = 68 ml

Pressure P2 = ?

FROM THE FORMULA OF GAS LAW:

=

volume is constant so, omitting volume values of temperature and pressure is put in the equation:

=

P2 =

P2 = 0.757 atm or 76.703 kPa is the pressure of the gas when cooled at -20 degrees.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3