Question

1. A sample of nitrogen gas at a pressure of 0.947 atm and a temperature of 213 °C, occupies a volume of 687 mL. If the gas is cooled at constant pressure until its volume is 570 mL, the temperature of the gas sample will be 2. A helium-filled weather balloon has a volume of 615 L at 23 °C and 754 mm Hg. It is released and rises to an altitude of 7.93 km, where the pressure is 323 mm Hg and the temperature is -29 °C. The volume of the balloon at this altitude is

Answer 1

1.)

we know the equation

P₁V₁/T₁ = P₂V₂/T₂ then

T₂ = P₂V₂T₁/P₁V₁

where, P₁ = initial pressure = 0.947 atm

V₁ = initial volume = 687 ml

T₁ = initial tempreture = 213⁰C

P₂ = final pressure = 0.947 atm

T₂ = final tempreture = ?

V₂ = final volume = 570 ml

substitute value in above equation

T₂ = 0.947 570 213 / 0.947 687 = 176.72⁰C

final tempreture of gas sample wll be 176.72⁰C

2.)

we know the equation

P₁V₁/T₁ = P₂V₂/T₂ then

V₂ = P₁V₁T₂/T₁P₂

where, P₁ = initial pressure = 754 mm Hg

V₁ = initial volume = 615 L

T₁ = initial tempreture = 23⁰C = 23+273 = 296 K

P₂ = final pressure = 323 mm Hg

T₂ = final tempreture = -29⁰C = -29+273 = 244 K

V₂ = final volume = ?

substitute value in above equation

V₂ = 754 615 244 / 296 323 = 1183.43 L

final volume of gas = 1183.43

(note - in this example temprature converted to kelvin only for to avoid confusion in negative sign in calculation otherwise no need to convert)