A student wanted to apply his knowledge of thermodynamics to some practical applications. He performed a number of experiments. In an experiment, he took a closed vessel in which he dissolved 28 g of iron in hydrochloric acid at 27^oC. In another experiment, he dissolved the same amount of iron but in open vessel. However, he compressed the gas evolved to 10 atm pressure at 27^oC and then again allowed to expand it isothermally and reversibly until the pressure fell down to 1 atm. In one case, he used the work obtained to lift a heavy body mass 20 kg till the pressure fell down to atmospheric pressure and in another case (on repeating the expt.), he used the work obtained to heat up 1 litre of water.

a) In the first expt., calculate the work done by the system.

b) In the 2nd expt., calculate work done by the system if the gas were not condensed.

c) What is work done by compressed gas?

d) Calculate the height through which body would be lifted.

e) Calculate the rise in temperature.

Let's break down the experiments and calculations step by step. The student is exploring the principles of thermodynamics through practical applications involving iron and hydrochloric acid. We will analyze each part of the question systematically.

Work Done in the First Experiment

In the first experiment, the student dissolves 28 g of iron in hydrochloric acid in a closed vessel. The reaction produces hydrogen gas, which can be represented as:

• Fe + 2 HCl → FeCl₂ + H₂

To find the work done by the system, we can use the formula for work done by a gas during expansion:

W = -PΔV

However, since the vessel is closed and the gas does not expand against an external pressure, the work done in this case is zero:

W₁ = 0 J

Work Done in the Second Experiment

In the second experiment, the student dissolves the same amount of iron in an open vessel, allowing the gas to escape. However, the gas is then compressed to 10 atm and allowed to expand isothermally and reversibly to 1 atm. The work done by the gas during this expansion can be calculated using the formula for isothermal expansion:

W = nRT ln(V_f/V_i)

Where:

• n = number of moles of gas
• R = ideal gas constant (8.314 J/(mol·K))
• T = temperature in Kelvin (27°C = 300 K)
• V_f = final volume
• V_i = initial volume

To find n, we first calculate the number of moles of hydrogen produced:

n = mass / molar mass = 28 g / 56 g/mol = 0.5 mol

Next, we need to determine the initial and final volumes. Using the ideal gas law, we can find the volume at 10 atm:

P₁V₁ = nRT

At 10 atm, we can rearrange to find V₁:

V₁ = nRT / P₁ = (0.5 mol)(8.314 J/(mol·K))(300 K) / (10 atm × 101.325 kPa/atm) ≈ 0.123 m³

For the final volume at 1 atm:

V₂ = nRT / P₂ = (0.5 mol)(8.314 J/(mol·K))(300 K) / (1 atm × 101.325 kPa/atm) ≈ 0.0123 m³

Now we can calculate the work done:

W = nRT ln(V_f/V_i) = (0.5)(8.314)(300) ln(0.0123/0.123)

Calculating this gives:

W ≈ -1,500 J

Work Done by Compressed Gas

The work done by the compressed gas during the expansion from 10 atm to 1 atm is the same as calculated above, which is:

W = -1,500 J

Lifting the Heavy Body

To find the height through which the 20 kg mass can be lifted using the work done, we can use the formula:

W = mgh

Where:

• m = mass (20 kg)
• g = acceleration due to gravity (approximately 9.81 m/s²)
• h = height (unknown)

Rearranging gives:

h = W / (mg) = -1,500 J / (20 kg × 9.81 m/s²) ≈ -7.65 m

Since work done is negative in this context, it indicates that the energy is used to lift the mass, thus the height is approximately 7.65 m.

Calculating the Rise in Temperature

To find the rise in temperature when the work done is used to heat 1 liter of water, we can use the formula:

Q = mcΔT

Where:

• Q = heat energy (which is equal to the work done, 1,500 J)
• m = mass of water (1 liter = 1 kg)
• c = specific heat capacity of water (approximately 4,186 J/(kg·°C))
• ΔT = change in temperature (unknown)

Rearranging gives:

ΔT = Q / (mc) = 1,500 J / (1 kg × 4,186 J/(kg·°C)) ≈ 0.36 °C

In summary, the calculations yield:

• Work done in the first experiment: 0 J
• Work done in the second experiment: -1,500 J
• Height lifted: approximately 7.65 m
• Rise in temperature of water: approximately 0.36 °C

This exploration of thermodynamics through practical experiments illustrates the principles of energy transfer and work done in chemical reactions and gas behavior.