Question

PLEASE ANSWER ALL PARTS:

1a. A sample of neon gas at a pressure of 1.13 atm and a temperature of 20.6 °C, occupies a volume of 597 mL. If the gas is compressed at constant temperature until its pressure is 1.63 atm

The volume of the gas sample will be _______ mL.

1b. A sample of helium gas at a pressure of 1.11 atm and a temperature of 29.1 °C, occupies a volume of 17.4 liters. If the gas is allowed to expand at constant temperature to a volume of 27.5 liters.

The pressure of the gas sample will be______ atm.

1c. A sample of neon gas at a pressure of 0.934 atm and a temperature of 239 °C, occupies a volume of 661 mL. If the gas is cooled at constant pressure until its volume is 572 mL,

the temperature of the gas sample will be _______ °C.

1d. A sample of helium gas at a pressure of 933 mm Hg and a temperature of 72 °C, occupies a volume of 14.1 liters. If the gas is cooled at constant pressure to a temperature of 32 °C,

The volume of the gas sample will be_______ L.

Answer 1

a)

Given:

Vi = 597 mL

Pi = 1.13 atm

Pf = 1.63 atm

use:

Pi*Vi = Pf*Vf

1.13 atm * 597 mL = 1.63 atm * Vf

Vf = 414 mL

Answer: 414 mL

b)

Given:

Pi = 1.11 atm

Vi = 17.4 L

Vf = 27.5 L

use:

Pi*Vi = Pf*Vf

1.11 atm * 17.4 L = Pf * 27.5 L

Pf = 0.702 atm

Answer: 0.702 atm

c)

Given:

Ti = 239.0 oC

= (239.0+273) K

= 512 K

Vi = 661 mL

Vf = 572 mL

use:

Vi/Ti = Vf/Tf

661.0 mL / 512.0 K = 572.0 mL / Tf

Tf = 443 K

Tf = 170 oC

Answer: 170 oC

d)

Given:

Vi = 14.1 L

Ti = 72.0 oC

= (72.0+273) K

= 345 K

Tf = 32.0 oC

= (32.0+273) K

= 305 K

use:

Vi/Ti = Vf/Tf

14.1 L / 345.0 K = Vf / 305.0 K

Vf = 12.5 L

Answer: 12.5 L