Question

Part A

The pressure inside a hydrogen-filled container was 2.10 atm at 21 ∘C. What would the pressure be if the container was heated to 88 ∘C ?

Express your answer with the appropriate units.

Part B

At standard temperature and pressure (0 ∘C and 1.00 atm ), 1.00 mol of an ideal gas occupies a volume of 22.4 L. What volume would the same amount of gas occupy at the same pressure and 95 ∘C ?

Express your answer with the appropriate units.

Answer 1

Ans. # Part A. Using Ideal gas equation: PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Putting the values in above equation for initial conditions (2.10 atm, 21.0⁰C, volume constant)-

            2.10 atm x V = n x R x 294.15 K

            Or, n = 2.10 atm V / (294.15 RK)

Hence, n = 0.007139 atm V / RK

# Putting the values in equation 1 for (n, and 88.0⁰C, V constant)

            P x V = (0.007139 atm V / RK) x R x 361.15 K

            Or, P = 2.57824985 atm V/ V

            Hence, P = 2.58 atm

Therefore, pressure at 88.0⁰C = 2.58 atm

# Part B. Calculate the volume of 1mol ideal gas at 95.0⁰C using Charles’s Law as follow-

V1 / T1 (STP) = V2 / T2 (95.0⁰C)               - temperature must be in terms of K

Or, 22.4 L / 273.15K = V2 / 368.15K

Or, V2 = (22.4 L / 273.15K) x 368.15K

Hence, V2 = 30.19 L

Therefore, volume of 1.0 mol ideal gas at 95.0⁰C = 30.19 L