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If the temperature of 10 mole of ideal gas is changed from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) at constant pressure of \(2 \mathrm{~atm}\), then the work done in the process is \((R=8.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K})\) (a) \(8.3 \times 10^{-3} \mathrm{~J}\) (b) \(8.3 \times 10^{-2} \mathrm{~J}\) (c) \(8.3 \times 10^{4} \mathrm{~J}\) (d) \(8.3 \times 10^{3} \mathrm{~J}\)

Short Answer

Expert verified
The work done is \(8.3 \times 10^{3} \, \mathrm{J}\). Hence, the answer is (d).

Step by step solution

01

Understand the Question

We need to calculate the work done on an ideal gas when it undergoes a change in temperature from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) at constant pressure. The number of moles is 10, and the pressure is \(2 \, \mathrm{atm}\). The gas constant \(R\) is given as \(8.3 \, \mathrm{J}/\mathrm{mol}-\mathrm{K}\).
02

Convert Temperatures to Kelvin

First, convert the temperatures from Celsius to Kelvin because the ideal gas equation requires temperature in Kelvin. - Initial temperature in Kelvin: \(T_1 = 0^{\circ} \mathrm{C} + 273.15 = 273.15 \, \mathrm{K}\)- Final temperature in Kelvin: \(T_2 = 100^{\circ} \mathrm{C} + 273.15 = 373.15 \, \mathrm{K}\)
03

Use the Work Formula for Isobaric Process

The formula for work done during an isobaric (constant pressure) process is:\[ W = P \Delta V \]where \(\Delta V\) is the change in volume. But, using the ideal gas law, we can relate it to temperature change:\[ W = nR(T_2 - T_1) \]
04

Substitute Known Values

Substitute the known values into the work formula:\[ W = 10 \, \text{moles} \times 8.3 \, \mathrm{J/mol-K} \times (373.15 - 273.15) \, \mathrm{K} \]
05

Calculate the Work Done

Now, calculate the work done:\[ W = 10 \times 8.3 \times 100 = 8300 \, \mathrm{J} \]
06

Compare with Given Choices

The calculated work done matches choice (d): \(8.3 \times 10^{3} \, \mathrm{J}\). Thus, this is the correct answer.

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Key Concepts

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

Isobaric Process
In thermodynamics, an isobaric process is a transformation where the pressure remains constant. It derives its name from the Greek words 'iso' meaning equal, and 'baros' meaning pressure. The word itself translates to equal pressure.
During this process, any heat added or removed from the system results in changes in the system’s volume and temperature while the pressure remains unchanged.

When dealing with gases, an isobaric process is often visualized on a Pressure-Volume (P-V) diagram as a straight, horizontal line, showing that pressure does not change as the volume varies.
  • This concept is critical in processes like heating where the pressure is kept constant, resulting in a linear rise in volume when temperature increases.
  • It's essential in understanding real-life applications, like the expansion of gas in a piston at a constant pressure which is common in engines and refrigeration cycles.
It's important to know that during an isobaric process, the work done by or on the system can be quantified using the equation: \(W = P \Delta V\). However, for ideal gases, this is often expressed in terms of temperature as shown in the original problem.
Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry and physics that relates a gas’s pressure, volume, temperature, and the number of moles. It is often expressed as \(PV = nRT\), where:
  • \(P\) stands for pressure
  • \(V\) is the volume
  • \(n\) represents the number of moles
  • \(R\) is the ideal gas constant
  • \(T\) is the temperature in Kelvin
This relationship allows us to predict how a gas will behave under different conditions. In our context of an isobaric process, it helps relate changes in temperature directly to changes in volume when pressure and moles are constant.

The ideal gas law assumes the gas behaves perfectly and follows the equation precisely, which may not be true in real-life conditions where gases might deviate slightly. Nonetheless, it provides an invaluable approximation for understanding gas behavior under various thermodynamic processes, serving as a baseline for more complex models.
Temperature Conversion to Kelvin
Temperature conversion to Kelvin is a crucial step when dealing with gas laws, especially the ideal gas law. Kelvin is the SI unit of temperature and is always used in scientific equations involving gases.
Celsius and Kelvin have the same magnitude, but Kelvin starts at absolute zero, the coldest possible temperature. Therefore, to convert Celsius to Kelvin, use the formula:
  • \(T_{K} = T_{C} + 273.15\)
This conversion is necessary to maintain the coherence of units and to ensure accuracy of the laws and formulas used in calculations.

For example, in the original exercise, the given temperatures were 0°C and 100°C. Converting these to Kelvin:
  • The initial temperature \(T_1\) becomes 273.15 K
  • The final temperature \(T_2\) becomes 373.15 K
Understanding this conversion is vital for students to accurately apply formulas like the ideal gas law and calculate things such as work done or changes in volume in a controlled thermodynamic process.

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Most popular questions from this chapter

In changing the state of a gas adiabatically from an equilibrium state \(A\) to another equilibrium state \(B\), an amount of work equal to \(22.3 \mathrm{~J}\) is done on the system. If the gas is taken from state \(A\) to \(B\) via a process in which the net heat absorbed by the system is \(9.35\) cal, how much is the net work done by the system in the latter case? (a) \(15.6 \mathrm{~J}\) (b) \(11.2 \mathrm{~J}\) (c) \(14.9 \mathrm{~J}\) (d) \(16.9 \mathrm{~J}\)

For a gas the difference between the two specific heats is \(4150 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\). What is the specific heats at constant volume of gas it the ratio of the specific heat is \(1.4\) ? (a) \(8475 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\) (b) \(5186 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\) (c) \(1660 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\) (d) \(10375 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\)

A monoatomic ideal gas, initially at temperature \(T_{1}\) is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature \(T_{2}\) by releasing the piston suddenly. If \(L_{1}, L_{2}\) are the lengths of the gas column before and after expansion respectively, then \(T_{1} / T_{2}\) is given by (a) \(\left(L_{1} / L_{2}\right)^{2 / 3}\) (b) \(\left(L_{1} / L_{2}\right)\) (c) \(L_{1} / L_{2}\) (d) \(\left(L_{2} / L_{1}\right)^{2 / 3}\)

A Carnot's engine works between a source at a temperature of \(27^{\circ} \mathrm{C}\) and a sink at \(-123^{\circ} \mathrm{C}\). Its efficiency is (a) \(0.5\) (b) \(0.25\) (c) \(0.75\) (d) \(0.4\)

A gas at pressure \(6 \times 10^{5} \mathrm{Nm}^{-2}\) and volume \(1 \mathrm{~m}^{3}\) and its pressure falls to \(4 \times 10^{5} \mathrm{Nm}^{-2} .\) When its volume is \(3 \mathrm{~m}^{3}\).Given that the indicator diagram is a straight line, work done by the system is (a) \(6 \times 10^{5} \mathrm{~J}\) (b) \(3 \times 10^{5} \mathrm{~J}\) (c) \(4 \times 10^{5} \mathrm{~J}\) (d) \(10 \times 10^{5} \mathrm{~J}\)

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