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 to understand the thermodynamic principles at play. The student performed two experiments involving the dissolution of iron in hydrochloric acid, and we will analyze the work done in each case, as well as the height a mass can be lifted and the temperature rise of water.

First Experiment: Work Done by the System

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

• Fe + 2 HCl → FeCl2 + H2↑

From the balanced equation, we see that 1 mole of iron produces 1 mole of hydrogen gas. The molar mass of iron (Fe) is approximately 56 g/mol, so 28 g corresponds to:

Number of moles of iron: 28 g / 56 g/mol = 0.5 moles

Thus, the reaction produces 0.5 moles of hydrogen gas. At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters. Therefore, the volume of hydrogen gas produced is:

Volume of H2: 0.5 moles × 22.4 L/mol = 11.2 L

Since the vessel is closed, the work done (W) by the system can be calculated using the formula:

W = -PΔV

Here, P is the pressure (assuming atmospheric pressure, 1 atm = 101.3 kPa) and ΔV is the change in volume. However, since the gas does not expand against any external pressure in a closed vessel, the work done is zero:

W = 0 J

Second Experiment: Work Done by the System with Gas Expansion

In the second experiment, the student dissolved the same amount of iron in an open vessel and compressed the gas to 10 atm. When the gas expands isothermally and reversibly to 1 atm, we can calculate the work done using the formula:

W = -nRT ln(P2/P1)

Where:

• n = number of moles of gas (0.5 moles)
• R = universal gas constant (8.314 J/(mol·K))
• T = temperature in Kelvin (27°C = 300 K)
• P1 = initial pressure (10 atm)
• P2 = final pressure (1 atm)

Substituting the values:

W = -0.5 moles × 8.314 J/(mol·K) × 300 K × ln(1 atm / 10 atm)

Calculating the natural logarithm:

ln(1/10) = -2.302

Now substituting this back into the equation:

W = -0.5 × 8.314 × 300 × (-2.302)

W ≈ 2877.3 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 approximately:

W ≈ 2877.3 J

Height Through Which the Body Would Be Lifted

To find the height (h) through which a 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 the formula gives:

h = W / (mg)

Substituting the values:

h = 2877.3 J / (20 kg × 9.81 m/s²)

h ≈ 14.66 m

Temperature Rise in Water

For the second experiment, if the work done is used to heat 1 liter of water, we can calculate the rise in temperature (ΔT) using the formula:

Q = mcΔT

Where:

• Q = work done (2877.3 J)
• m = mass of water (1 liter = 1 kg)
• c = specific heat capacity of water (approximately 4.18 J/(g·°C) or 4180 J/(kg·°C))
• ΔT = temperature rise (unknown)

Rearranging gives:

ΔT = Q / (mc)

Substituting the values:

ΔT = 2877.3 J / (1 kg × 4180 J/(kg·°C))

ΔT ≈ 0.688 °C

Summary of Results

To summarize the findings from the experiments:

• Work done in the first experiment: 0 J
• Work done in the second experiment: 2877.3 J
• Height lifted: 14.66 m
• Temperature rise of water: 0.688 °C

This analysis illustrates the practical applications of thermodynamics in real-world scenarios, demonstrating how energy transformations can be harnessed for mechanical work and heating processes.