10 mole of ideal gas expand isothermally and reversibly from a pressure of \(10 \mathrm{~atm}\) to \(1 \mathrm{~atm}\) at \(300 \mathrm{~K}\). What is the largest mass which can lifted through a height of 100 meter? (a) \(31842 \mathrm{~kg}\) (b) \(58.55 \mathrm{~kg}\) (c) \(342.58 \mathrm{~kg}\) (d) None of these

Short Answer

Expert verified
The largest mass which can be lifted through a height of 100 meters is none of the provided options (d) None of these.

Step by step solution

01

Calculate the work done by the gas

For an isothermal reversible expansion of an ideal gas, the work done can be calculated using the formula: \(W = nRT \ln\left(\frac{V_f}{V_i}\right)\) where \(n\) is the number of moles, \(R\) is the gas constant, \(T\) is the temperature, \(V_f\) is the final volume, and \(V_i\) is the initial volume. The initial and final volumes can also be represented in terms of the initial and final pressures (\(P_i\) and \(P_f\)) for an isothermal process using Boyle's Law (\(P_iV_i = P_fV_f\)). Here we start with \(n = 10\) moles, \(R = 0.0821\) L\(\cdot\)atm/K\(\cdot\)mol, \(T = 300\) K, \(P_i = 10\) atm, and \(P_f = 1\) atm. The work done by the gas, \(W\), can be found as follows: \(W = 10 \times 0.0821 \times 300 \ln\left(\frac{1}{10}\right)\).
02

Calculate the work in joules

The calculated work will be in liter-atmosphere units. To convert liter-atmosphere to joules, use the conversion factor \(1 \mathrm{L}\cdot\mathrm{atm} = 101.325 \mathrm{J}\). Multiplying the work by this conversion factor gives the work done by the gas in joules.
03

Relate the work done to lifting a mass

The work done by the gas can be equated to the work required to lift a mass through a height in the earth's gravitational field. The work needed to lift a mass \(m\) through a height \(h\) is given by \(W = mgh\), where \(g\) is the acceleration due to gravity (approximated as \(9.81 \mathrm{m/s^2}\)). Rearranging the formula to solve for mass, we get \(m = \frac{W}{gh}\).
04

Calculate the mass that can be lifted

Substitute the work calculated in joules and the given height of 100 meters into the formula to find the mass that can be lifted by the work done during the expansion.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Done by Gas
The concept of work done by a gas is central to understanding several physical processes, including the operation of heat engines and refrigerators. During expansion, a gas can do work against an external pressure. This work is essentially the energy transferred by the gas to its surroundings. In thermodynamics, the work done by a gas during an expansion or compression is a critical parameter.

In an isothermal reversible expansion, such as the exercise mentioned, the temperature of the gas remains constant, which implies that the internal energy of the gas doesn't change, and the work done is equal to the heat absorbed by the system from the surroundings. The formula for the work done in an isothermal process utilizes the natural logarithm of the volume ratio and is given by:
\[W = nRT \times \text{ln}\left(\frac{V_f}{V_i}\right)\]
where \(n\) is the number of moles of gas, \(R\) is the ideal gas constant, \(T\) is the absolute temperature, \(V_f\) is the final volume, and \(V_i\) is the initial volume. For an ideal gas undergoing an isothermal expansion, this formula allows us to calculate the work done by the gas as it expands against an external pressure.
Ideal Gas Law
The Ideal Gas Law is a fundamental principle in chemistry and physics that describes the behavior of an ideal gas. An ideal gas is a hypothetical gas that perfectly follows the relationship between pressure, volume, temperature, and the number of moles of gas. Although no real gas is truly 'ideal', many come close under a range of conditions, making the ideal gas law very useful. The law is mathematically represented as:
\[PV = nRT\]
where \(P\) stands for pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the absolute temperature in Kelvin. This law combines several simpler gas laws, such as Boyle's Law, Charles's Law, and Avogadro's Law, into one comprehensive equation. In the context of the exercise provided, the ideal gas law helps to connect the initial and final states of the gas during its isothermal expansion.
Boyle's Law
Boyle's Law is a gas law which states that the pressure and volume of a gas are inversely proportional to each other at a constant temperature for a fixed mass of an ideal gas. In other words, if the volume of a gas increases, the pressure decreases, provided the temperature and the number of moles of gas remain constant, and vice versa. This relationship can be mathematically expressed as:
\[P_iV_i = P_fV_f\]
where \(P_i\) and \(P_f\) are the initial and final pressures, and \(V_i\) and \(V_f\) are the initial and final volumes of the gas, respectively. Boyle's Law is particularly relevant in the exercise because it allows us to relate the initial state to the final state of the system during the isothermal process, which is necessary to calculate the work done by the expanding gas.
Gravitational Potential Energy
Gravitational potential energy is the energy stored in an object due to its position in a gravitational field. More simply, it is the energy that an object has because of its height above the ground. When you lift an object against the force of gravity, you’re increasing its gravitational potential energy. The formula to calculate this energy for an object of mass \(m\) lifted to a height \(h\) in the Earth's gravity is given by:
\[E_p = mgh\]
where \(g\) represents the acceleration due to gravity, which is approximately \(9.81 \text{ m/s}^2\) on Earth. This concept is especially useful in relating the work done by the gas in the provided exercise to the physical task of lifting a mass through a certain height. Through this comparison, students can better visualize the amount of work that the gas can perform as a tangible measure, like the weight of an object being lifted, enriching their understanding of energy conversion and conservation principles.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The enthalpy of the reaction forming \(\mathrm{PbO}\) according to the following equation is \(438 \mathrm{~kJ}\). What heat energy (kJ) is released in formation of \(22.3 \mathrm{~g} \mathrm{PbO}(s)\) ? (Atomic weights : \(\mathrm{Pb}=207, \mathrm{O}=16.0\) ) $$ 2 \mathrm{~Pb}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s) $$ (a) \(21.9\) (b) \(28.7\) (c) \(14.6\) (d) \(34.2\)

Chloroform has \(\Delta H_{\text {vaporization }}=29.2 \mathrm{~kJ} / \mathrm{mol}\) and boils at \(61.2^{\circ} \mathrm{C} .\) What is the value of \(\Delta S_{\text {vaporization }}\) for chloroform? (a) \(87.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) (b) \(477.1 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) (c) \(-87.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) (d) \(-477.1 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\)

For a reversible adiabatic ideal gas expansion \(\frac{\mathrm{a} P}{P}\) is equal to : (a) \(\gamma \frac{d V}{V}\) (b) \(-\gamma \frac{d V}{V}\) (c) \(\left(\frac{\gamma}{\gamma-1}\right) \frac{d V}{V}\) (d) \(\frac{d V}{V}\)

A \(0.05 \mathrm{~L}\) sample of \(0.2 \mathrm{M}\) aqueous hydrochloric acid is added to \(0.05 \mathrm{~L}\) of \(0.2 \mathrm{M}\) aqueous ammonia in a calorimeter. Heat capacity of entire calorimeter system is \(480 \mathrm{~J} / \mathrm{K}\). The temperature increase is \(1.09 \mathrm{~K}\). Calculate \(\Delta_{r} H^{\circ}\) in \(\mathrm{kJ} / \mathrm{mol}\) for the following reaction: $$ \mathrm{HCl}(a q .)+\mathrm{NH}_{3}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ (a) \(-52.32\) (b) \(-61.1\) (c) \(-55.8\) (d) \(-58.2\)

Which of the following exprésstons' is 'true for an ideal gas ? (a) \(\left(\frac{\partial V}{\partial T}\right)_{P}=0\) (b) \(\left(\frac{\partial P}{\partial T}\right)_{V: n}=0^{6}\) (c) \(\left(\frac{\partial U}{\partial V}\right)_{T}=0\) (d) \(\left(\frac{\partial U}{\partial T}\right)_{V}=0\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free