Question

33. 2.50 L of So, gas originally at 45.0 degrees Celcius is warmed to 85.0 degrees Celcius. What is the new Volume of the sulfur dioxide gas if the pressure remains constant? 

34. A birthday party balloon is filled with 120 mL of Helium gas at a pressure of 0.750 atm pressure and a temperature of 22.0 degrees Celcius. What will be the new Volume if the pressure is increased to 1.25 atm and the temperature is raised to 70.0 degrees Celcius?

Answer 1

Unless it is mentioned in the question that the given gas exhibits real gas properties, we assume it to be ideal and use ideal gas equations and ideal gas laws to solve the numerical problems.

Solution: Given,

Volume of SO2 gas at 45 degree C { 45 + 273.15 = 318.15 K } = 2.50 L .

To find: Volume of SO2 gas at 85 degree C { 273.15 + 85 = 358.15 K }.

We know the ideal gas equation, PV = nRT

where P is the pressure of the gas

V is the volume of the gas

n is the number of moles of the gas

R is the gas constant

T is the temperature in Kelvin.

If we name the initial conditions as P1, V1 and T1 and final conditions as P2, V2 and T2 , then we have

P1V1 = nRT1 .....(1)

P2V2 = nRT2 ....(2) { no. of moles are same in both the cases because only the temperature has increased }

Divide equation 1 by 2,

P1V1/P2V2 = nRT1 / nRT2

Since P1 = P2 { As mentioned in the problem that the pressure is constant }

V1/V2 = T1/T2

or V2 = V1T2/T1

or V2 = 2.50 L * 358.15 / 318.15

or V2 = 2.81 L Ans.

Solution 2: Given,

Initial conditions,

P1 = 0.750 atm

V1 = 120 ml

T1 = 22 degree Celsius = 22 + 273.15 = 295.15 K

Final conditions,

P2 = 1.25 atm

T2 = 70 degree Celsius = 343.15 K

V2 = ?

Using the ideal gas equation in both the conditions we have,

P1V1 = nRT1 .....(1)

P2V2 = nRT2 .....(2) { no. of moles are same in both the conditions }

Divide equation 1 by equation 2,

P1V1/P2V2 = nRT1/nRT2

P1V1/P2V2 = T1/T2

On simplification, we have,

V2 = P1V1T2 / P2T1

V2 = { 0.750 atm * 120 ml * 295.15 K } / { 1.25 atm * 343.15 K }

or V2 = 61.93 ml Ans.