A sample of argon of mass 10.4 g occupies 2100 ml at 30 oC .
(i) Calculate the work done by the gas when it expands isothermally against a constant external pressure of 31.1kPa until its volume has increased by 483.1ml.
(ii) Calculate the work done by the gas if the same expansion occurs reversibly.
(iii) What is the change in internal energy for the expansion process in (i)?
(iv) For the reversible expansion process, what are the values of DU, DH and DS ?
To solve the given problems, we will be using the ideal gas law equation and its derived equations for the required calculations. Let's proceed step by step.
Step 1: Convert the given temperature to Kelvin.
The given temperature is 30°C. To convert it to Kelvin, we use the formula:
Kelvin = Celsius + 273.15
Therefore, the temperature in Kelvin = 30 + 273.15 = 303.15 K
Step 2: Calculate the initial and final volumes.
The initial volume is given as 2100 mL, and the final volume is obtained by adding 483.1 mL.
Therefore, the initial volume = 2100 mL = 2.1 L
And the final volume = 2100 mL + 483.1 mL = 2583.1 mL = 2.5831 L
Step 3: Calculate the number of moles.
The number of moles can be calculated using the formula:
Moles = mass / molar mass
The molar mass of argon is approximately 39.95 g/mol.
Therefore, the number of moles = 10.4 g / 39.95 g/mol ≈ 0.260 moles
Step 4: Solve each part of the question.
(i) Calculate the work done by the gas when it expands isothermally against a constant external pressure of 31.1 kPa until its volume has increased by 483.1 mL.
For an isothermal process, the work done by the gas can be calculated as:
Work = -Pext × ΔV
Where Pext is the external pressure, and ΔV is the change in volume.
Convert the external pressure to atm:
31.1 kPa = 31.1 / 101.325 atm ≈ 0.3064 atm
Calculate the work:
Work = -0.3064 atm × (2.5831 L - 2.1 L)
Work ≈ -0.3064 atm × 0.4831 L ≈ -0.1481 L atm
(ii) Calculate the work done by the gas if the same expansion occurs reversibly.
For a reversible expansion, the work done can be calculated using the equation:
Work reversible = -nRT ln(Vf / Vi)
Where n is the number of moles, R is the ideal gas constant (0.0821 L atm / K mol), T is the temperature in Kelvin, and ln represents the natural logarithm.
Calculate the work reversible:
Work reversible = -0.260 mol × 0.0821 L atm / K mol × 303.15 K × ln(2.5831 L / 2.1 L)
(iii) What is the change in internal energy for the expansion process in (i)?
For an isothermal process, the change in internal energy (ΔU) is zero.
(iv) For the reversible expansion process, what are the values of ΔU, ΔH, and ΔS?
For a reversible expansion, the change in internal energy (ΔU) can be calculated using the equation:
ΔU = q + W
Where q is the heat transferred and W is the work done.
Since the expansion is isothermal, the temperature remains constant, so ΔU = 0.
The enthalpy change (ΔH) for an ideal gas is given by:
ΔH = ΔU + PΔV
For a reversible expansion, ΔU = 0, so:
ΔH = PΔV
The change in entropy (ΔS) can be calculated using the equation:
ΔS = q / T
For an isothermal process, ΔS = q / T = ΔH / T
Using the values calculated earlier, we can determine ΔH and ΔS for the given reversible expansion.