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In a cylinder, 1.20 mol of an ideal monatomic gas, initially at 3.60 × 105 Pa and 300 K, expands until its volume triples. Compute the work done by the gas if the expansion is (a) isothermal; (b) adiabatic; (c) isobaric. (d) Show each process in a pV-diagram. In which case is the absolute value of the work done by the gas greatest? Least? (e) In which case is the absolute value of the heat transfer greatest? Least? (f) In which case is the absolute value of the change in internal energy of the gas greatest? Least?

Short Answer

Expert verified
Greatest work: Isothermal; Least work: Adiabatic. Greatest heat transfer: Isothermal; Least: Adiabatic. Greatest internal energy change: Adiabatic; Least: Isothermal.

Step by step solution

01

Identify Given Information

We have 1.20 mol of an ideal monatomic gas initially at pressure P1=3.60×105 Pa and temperature T1=300 K. The gas expands to triple its initial volume. We need to calculate the work done for three processes: isothermal, adiabatic, and isobaric.
02

Step 2(a): Calculate Work for Isothermal Expansion

In an isothermal process, the work done by the gas is given by W=nRTln(V2V1). Since the volume triples, V2V1=3. Using the ideal gas constant R=8.314 J/mol·K, and n=1.20, T=300, substitute these values:W=1.20×8.314×300×ln(3)Calculate W2.73×103 J.
03

Step 2(b): Calculate Work for Adiabatic Expansion

In an adiabatic process, the relationship PVγ=constant holds, where γ=53 for a monatomic ideal gas. Use V2V1=3 and the initial conditions to find work:W=P1V1P2V2γ1.First, find P2 using P1V1γ=P2V2γ then substitute to find the work. This involves solving:P2=P1(V1V2)γ=3.60×105×(13)5/3.Calculate the work W1.60×103 J.
04

Step 2(c): Calculate Work for Isobaric Expansion

In an isobaric process, the pressure remains constant. The work done is given by W=PΔV. Since the volume triples, ΔV=2V1. From the ideal gas law, V1=nRTP. Substitute P=3.60×105,V1=1.20×8.314×3003.60×105.Then, W=3.60×105×2×V1. Calculate W1.99×103 J.
05

Compare Work Done

Compare the calculated work done in each process:- Isothermal: W2.73×103 J (greatest)- Isobaric: W1.99×103 J- Adiabatic: W1.60×103 J (least)The greatest work done is during the isothermal process.
06

Compare Heat Transfer and Change in Internal Energy

For heat transfer: - In an isothermal process, heat transfer is maximized. - In an adiabatic process, there is no heat transfer. For change in internal energy: - In an isothermal process, internal energy does not change. - In an adiabatic process, the internal energy change is largest compared to isobaric and minimal in isothermal.

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Key Concepts

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

Isothermal Expansion
In an isothermal expansion, the temperature of an ideal gas remains constant throughout the process. This implies that even as the gas expands and does work on its surroundings, it must absorb heat from the environment to maintain a constant temperature. The work done by the gas can be calculated using the formula:
  • W=nRTln(V2V1)
Here, n is the amount of substance in moles, R is the ideal gas constant, T is the temperature in Kelvin, and V2/V1 is the volume ratio. During an isothermal process, because there is no change in internal energy, the heat added to the system equals the work done by the gas.
This makes isothermal processes unique as it means the system exchanges the most heat with its surroundings among all types of thermodynamic processes.
Adiabatic Expansion
An adiabatic expansion occurs when gas expands without exchanging heat with its surroundings. During this process, all the work done by the gas causes a drop in its internal energy, resulting in a decrease in temperature. The adiabatic condition follows the relationship:
  • PVγ=constant
where γ=CpCv is the heat capacity ratio. For an ideal monatomic gas, γ is 5/3.
Adiabatic processes can be fast because there's no concern about the heat exchange.
This process shows the greatest change in temperature and internal energy, balancing the reduced work done compared to isothermal expansion.
Isobaric Process
An isobaric process occurs under constant pressure. In this scenario, when a gas expands, the work done is simply the product of the constant pressure and the change in volume. This can be expressed as:
  • W=PΔV
where ΔV is the change in volume, calculated as the difference between final and initial volumes. This type of process is frequently encountered in everyday scenarios, such as heating air in a piston at constant pressure.
Though the amount of work done by the gas is more than in an adiabatic expansion, it is still less than the work done during an isothermal expansion when all other factors remain constant.
Work Done by Gas
Work done by gas in any thermodynamic process can be understood as the energy transferred by the system to its surroundings due to changes in volume. For ideal gases, understanding how different processes affect work done is crucial:
  • Isothermal: Maximum work done as pressure decreases slowly over a large distance because temperature remains constant.
  • Adiabatic: Least work done because of no heat exchange and subsequent rapid volume change.
  • Isobaric: Intermediate work done, as pressure stays the same throughout the volume change.
The concept of work done is key in determining the efficiency of mechanical engines, refrigeration cycles, and various other applications in thermodynamics.
Heat Transfer Calculations
Heat transfer is a crucial concept that defines how energy moves in and out of a gas during thermodynamic processes. It varies with each type of process:
  • Isothermal: Heat transfer is non-zero and equals the work done, making it substantial. This is because temperature remains constant and heat compensates for the work done by the gas.
  • Adiabatic: There is no heat transfer. All energy changes occur internally through work done by or on the gas.
  • Isobaric: The heat transfer depends on both work done and changes in internal energy. Due to constant pressure, heat must be added or removed to change temperature.
Understanding heat transfer helps in designing systems where temperature control is essential, such as engines, and maintaining energy efficiency.

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Most popular questions from this chapter

You have a cylinder that contains 500 L of the gas mixture pressurized to 2000 psi (gauge pressure). A regulator sets the gas flow to deliver 8.2 L/min at atmospheric pressure. Assume that this flow is slow enough that the expansion is isothermal and the gases remain mixed. How much time will it take to empty the cylinder? (a) 1 h; (b) 33 h; (c) 57 h; (d) 140 h.

A cylinder with a piston contains 0.150 mol of nitrogen at 1.80 × 105 Pa and 300 K. The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure. (a) Show the series of processes in a pV-diagram. (b) Compute the temperatures at the beginning and end of the adiabatic expansion. (c) Compute the minimum pressure.

A cylinder with a piston contains 0.250 mol of oxygen at 2.40 × 105 Pa and 355 K. The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure. (a) Show the series of processes on a pV-diagram. Compute (b) the temperature during the isothermal compression; (c) the maximum pressure; (d) the total work done by the piston on the gas during the series of processes.

On a warm summer day, a large mass of air (atmospheric pressure 1.01 × 105 Pa) is heated by the ground to 26.0C and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why?) Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 0.850 × 105 Pa. Assume that air is an ideal gas, with Υ = 1.40. (This rate of cooling for dry, rising air, corresponding to roughly 1 C per 100 m of altitude, is called the dry adiabatic lapse rate.)

Two moles of an ideal monatomic gas go through the cycle abc. For the complete cycle, 800 J of heat flows out of the gas. Process ab is at constant pressure, and process bc is at constant volume. States a and b have temperatures Ta = 200 K and Tb = 300 K. (a) Sketch the pV-diagram for the cycle. (b) What is the work W for the process ca?

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