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A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at 1.00 × 105Pa and occupies a volume of 2.50 × 103 m3. (a) Find the initial temperature of the gas in kelvins. (b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.

Short Answer

Expert verified
(a) Initial temperature: 300.5 K. (b) Final conditions: (i) Isothermal: 300.5 K, 0.50 × 10⁵ Pa; (ii) Isobaric: 601.0 K, 1.00 × 10⁵ Pa; (iii) Adiabatic: 238.7 K, 0.316 × 10⁵ Pa.

Step by step solution

01

Understand Ideal Gas Law

The Ideal Gas Law is given by the equation PV=nRT. Here, P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin.
02

Solve for Initial Temperature

To find the initial temperature Ti, rearrange the ideal gas law: Ti=PVnR. Plug in the given values: P=1.00×105 Pa, V=2.50×103 m3, n=0.100 mol, and R=8.314 J/(mol·K), to get Ti=(1.00×105)(2.50×103)0.1008.314.
03

Calculate Initial Temperature

Perform the calculations: Ti=2.5×1020.8314300.5 K.
04

Solve Isothermal Expansion (i)

For isothermal expansion, the temperature remains constant. Therefore, Tf=Ti=300.5 K. Using PiVi=PfVf, rearrange to find Pf: Pf=PiViVf. Substitute Vf=2×2.50×103=5.00×103 m3 to get Pf=1.00×1052.50×1035.00×103=0.50×105 Pa.
05

Solve Isobaric Expansion (ii)

For an isobaric process, the pressure remains constant. So Pf=Pi=1.00×105 Pa. Using the ideal gas law TfTi=VfVi, calculate Tf=TiVfVi=300.55.00×1032.50×103=601.0 K.
06

Solve Adiabatic Expansion (iii)

For a monatomic gas, the adiabatic condition is given by PVγ=constant, where γ=53. Using Tf=Ti(ViVf)γ1, calculate Tf=300.5(2.50×1035.00×103)2/3238.7 K. For Pf, use Pf=Pi(ViVf)γ0.316×105 Pa.

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Key Concepts

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

Isothermal Process
In an isothermal process, the temperature remains constant throughout the expansion or compression of the gas. This means, that for an isothermal expansion, the initial and final temperatures are equal. According to the ideal gas law, when temperature is constant,
  • The product of pressure (P) and volume (V) remains constant: PV=extconstant.
  • For the expansion of the gas to twice its initial volume, Vf=2Vi.
  • Using this property, we can find the final pressure:Pf=PiViVf.
The most important aspect of the isothermal process is the constancy of temperature; hence, internal energy changes only through work done by or on the gas. This process requires perfect thermal equilibrium with the surroundings to keep the temperature unchanged.
Isobaric Process
In an isobaric process, the pressure remains constant. This phenomenon affects other gas properties as they change according to the ideal gas law. Since the pressure does not change:
  • The relationship between temperature and volume becomes directly proportional.
  • Using the equation TfTi=VfVi, it shows that the final temperature is determined by the change in volume.
For the gas to expand to twice its initial volume, the ideal gas law simplifies the calculation for the final state properties given constant pressure. During such processes, thermal energy is absorbed or released to maintain constant pressure, leading to a change in the gas's volume and temperature.
Adiabatic Expansion
Adiabatic processes are distinctive because they occur without the transfer of heat between the system and its surroundings. For a monatomic ideal gas, the key equations include:
  • The adiabatic equation: PVγ=extconstant, where γ=53 (ratio of specific heats for a monatomic gas).
  • Temperature changes even though no heat enters or leaves the system.
  • Final temperature can be determined with Tf=Ti(ViVf)γ1.
Since adiabatic processes occur without heat exchange, changes in internal energy are driven solely by work done on or by the gas. This process is significant in understanding how temperature and pressure can drastically change together in closed systems.

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Most popular questions from this chapter

Nitrogen gas in an expandable container is cooled from 50.0C to 10.0C with the pressure held constant at 3.00 × 105 Pa. The total heat liberated by the gas is 2.50 × 104 J. Assume that the gas may be treated as ideal. Find (a) the number of moles of gas; (b) the change in internal energy of the gas; (c) the work done by the gas. (d) How much heat would be liberated by the gas for the same temperature change if the volume were constant?

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