Temperature of one mole of an ideal gas is increased by one degree at constant pressure. Work done by the gas is (a) \(R\) (b) \(2 R\) (c) \(R / 2\) (d) \(3 R\)

Short Answer

Expert verified
R

Step by step solution

01

Understanding the Relationship between Pressure, Volume and Temperature for an Ideal Gas

For an ideal gas undergoing a process at constant pressure, the work done on or by the gas can be determined by the product of pressure and the change in volume. According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, T is the temperature, n is the number of moles, and R is the ideal gas constant.
02

Determining the Change in Volume

The change in temperature is 1 degree. We can express the initial state of the gas by the equation P*V1 = n*R*T1 and the final state as P*V2 = n*R*(T1+1). To find the change in volume, subtract the initial volume V1 from the final volume V2.
03

Calculating the Work Done

The work done at constant pressure is given by W = P * (V2 - V1). Using the ideal gas law, we can find V2 - V1 by substituting P*V1 = n*R*T1 and P*V2 = n*R*(T1+1), which simplifies to W = P * (n*R*T1 + n*R*1 - n*R*T1) = P * n*R*1. Since the pressure is constant and we have one mole of gas (n = 1), the work done is W = P * R * 1 = R.

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.

The Ideal Gas Law
The Ideal Gas Law is a cornerstone of understanding how gases behave under various conditions. Simply stated, it relates the pressure (P), volume (V), and temperature (T) of an ideal gas to the number of moles (n) and the ideal gas constant (R) through the equation PV = nRT. This equation assumes that the gas particles are in constant, random motion, and interact with each other and the container walls only through elastic collisions.

For a given amount of gas, if we change one of the variables, at least one other variable must also change to satisfy the equation. This relationship is crucial in calculating the work done by an ideal gas when it undergoes a change in state, such as a temperature increase, while keeping the pressure constant.
Thermodynamics and Work
In the realm of thermodynamics, work is a transfer of energy that results from a process, such as the expansion or compression of a gas. For an ideal gas, the work done during expansion or contraction can be calculated using the formula W = P * \(V2 - V1\), where W is the work done, P is the constant pressure, V1 is the initial volume, and V2 is the final volume.

It's important to remember that work is a path function, meaning its value depends on the specific process—how a system goes from one state to another. The idea of work connects thermodynamics with mechanical processes, providing a bridge between the microscopic interactions of particles and macroscopic observable phenomena.
Constant Pressure Process
A constant pressure process, often called an isobaric process, is a transformation of a system in which the pressure remains unchanged. When dealing with gases, a constant pressure process means that the gas may change volume and temperature while the external pressure remains steady.

This situation simplifies calculations since the work done by the gas is directly proportional to the change in volume, as expressed in the formula W = P \((V2 - V1)\). In isobaric processes, the area under the pressure-volume (P-V) curve graph represents the work done by or on the gas. Constant pressure processes are very common in everyday applications, such as inflating a balloon or operating a piston.
Volume and Temperature Relationship
For an ideal gas undergoing a process with constant pressure, there's a direct relationship between volume and temperature known as Charles's Law. This law states that if the pressure is held constant, the volume of a gas is directly proportional to its absolute temperature (measured in Kelvin).

Mathematically, it can be represented as V1/T1 = V2/T2, where V1 and V2 are the initial and final volumes, respectively, and T1 and T2 are the initial and final temperatures. This means if the temperature of the gas rises, so does its volume, and conversely, if the temperature decreases, the volume diminishes, provided the pressure remains unchanged.

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

A system is said to be in thermodynamic equilibrium with surrounding if (a) it is only in thermal equilibrium with surrounding. (b) it is in both thermal and chemical equilibrium with surrounding. (c) it is in thermal, chemical as well as mechanical equilibrium with the surrounding. (d) it is in thermal and mechanical equilibrium, but not in chemical equilibrium with surrounding.

The pressure and density of a diatomic gas \((\gamma=7 / 5)\) change from \(\left(P_{1}, d_{1}\right)\) to \(\left(P_{2},\right.\), \(d_{2}\) ) adiabatically. If \(d_{2} / d_{1}=32\), then what is the value of \(P_{2} / P_{1}\) ? (a) 32 (b) 64 (c) 128 (d) 256

The internal energy change when a system goes from state \(A\) to \(B\) is \(40 \mathrm{~kJ} /\) mol If the system goes from \(A\) to \(B\) by a reversible path and returns to state \(\mathrm{A}\) by an irreversible path, what would be the net change in internal energy? (a) \(40 \mathrm{~kJ}\) (b) \(>40 \mathrm{~kJ}\) (c) \(\leq 40 \mathrm{~kJ}\) (d) Zero

When one mole of an ideal gas is compressed to half of its initial volume and simultaneously heated to twice its temperature, the change in entropy is (a) \(C_{\mathrm{V}} \ln 2\) (b) \(C_{\mathrm{P}} \ln 2\) (c) \(R \ln 2\) (d) \(\left(C_{\mathrm{V}}-R\right) \ln 2\)

One mole of an ideal gas undergoes the following cyclic process: (i) Isochoric heating from \(\left(P_{1}, V_{1}, T_{1}\right)\) to double temperature. (ii) Isobaric expansion to double volume. (iii) Linear expansion (on \(P-V\) curve) to \(\left(P_{1}, 8 V_{1}\right) .\) (iv) Isobaric compression to initial state. If \(T_{1}=300 \mathrm{~K}\), the magnitude of net work done by the gas in the cyclic process is (a) \(2400 \mathrm{cal}\) (b) \(1200 \mathrm{cal}\) (c) \(4800 \mathrm{cal}\) (d) \(3600 \mathrm{cal}\)

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