Question

(Q.11)A gas has a volume of 2.8 L at a temperature of 27 o C. What temperature (o C) is needed to expand the volume to 15 L (P and n are held constant).

(Q.12)  Combined gas law problem: A balloon is filed with 500 mL of helium at a temperature of 27 o C and 755 mm Hg. As the balloon rises in the atmosphere, the pressure and temperature drop. What volume will it have when it reaches an altitude where the temperature is -33 o C and the pressure is 0.65 atm?

Please show all steps.

Answer 1

Q11 )

Inital volume of gas = 2.8 L = V1

Initial temperature of gas = T1 = 27 C = 300 K

final volume of gas = V2 = 15 L

Final temperature = T2 = ?

We will use ideal gas equation to solve the problem

PV = nRT

V / T = nR/P

When n and P are constant

V1/T1 = V2/T2

2.8 / 300 = 15 / T2

T2 = 1607.71 Kelvin

so the gas need to be heated to this temperature

Q12 ) Balloon filled with helium

volume of helium = 500mL = 0.5 L = V1

Temperature = 27 C = 273 + 27 K = 300 K = T1

Pressure = 755 mmHg = P1

760 mmHg = 1 atm

So 755 mmHg = 755 /760 atm = 0.99 atm

When it reaches an altitude,

Temperature = -33 C = 273-33 = 240 K = T2

Pressure = 0.65 atm = P2

Final volume = V2 = ?

We will use ideal gas equation to solve the problem,

PV = nRT

PV/ T = nR

For same number of moles of gas

P1V1 / T1 = P2V2/ T2

0.99 x 0.5 / 300 = 0.65 X V2 / 240

V2 = 0.00165 X 240 / 0.65 = 0.609 L or 609mL