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In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas release \(20 \mathrm{~J}\) of heat and \(8 \mathrm{~J}\) of work has done on the gas-If the initial internal energy of the gas was \(30 \mathrm{j}\), then the final internal energy will be (A) \(58 \mathrm{~J}\) (B) \(2 \mathrm{~J}\) (C) \(42 \mathrm{~J}\) (D) \(18 \mathrm{~J}\)

Short Answer

Expert verified
The final internal energy of the gas is \(18 \mathrm{~J}\), which is option (D).

Step by step solution

01

Write down the given values

We are given the following values: Initial internal energy (U鈧) = 30 J Heat released (Q) = -20 J (Since heat is released, it is negative) Work done on the gas (W) = 8 J (Since work is done on the gas, it is positive)
02

Apply the first law of thermodynamics

Using the first law of thermodynamics, we can find the change in internal energy (螖U): 螖U = Q + W
03

Substitute the given values

We can now substitute the given values into the equation: 螖U = (-20 J) + (8 J)
04

Calculate the change in internal energy

Now, we can calculate the change in internal energy: 螖U = -12 J
05

Find the final internal energy

To find the final internal energy (U鈧), we will use the equation: U鈧 = U鈧 + 螖U Substitute the given values: U鈧 = (30 J) + (-12 J)
06

Calculate the final internal energy

Finally, calculate the final internal energy: U鈧 = 18 J The final internal energy (U鈧) is 18 J, which corresponds to option (D).

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Key Concepts

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

internal energy
In the world of thermodynamics, the concept of internal energy is pivotal. Internal energy refers to the total energy contained within a system due to the movement of molecules and the forces between them. It is a state function, meaning it depends only on the current state of the system, not on how the system arrived at this state.

The unit for measuring internal energy is Joules (J), and any change in a system's internal energy signifies alterations in the system's microscopic properties. Consider a gas in a container: its internal energy includes the kinetic energy from the motion of its particles.

When energy enters or leaves a system in the form of heat or work, the internal energy changes. Here's a simple formula to express this:
  • Change in internal energy, \( \Delta U = U_2 - U_1 \)
  • where \( U_1 \) is the initial internal energy, and \( U_2 \) is the final internal energy.
The internal energy will increase if work is done on the system or heat is added. Conversely, it decreases if the system does work or loses heat.
heat transfer
Heat transfer is a fundamental concept in thermodynamics, describing the movement of thermal energy from a hotter object to a cooler one. In the context of the first law of thermodynamics, heat transfer plays a critical role in changing the energy state of a system.

Heat is denoted by \( Q \) and, in thermodynamic equations, it can be positive or negative:
  • Positive \( Q \): Heat is added to the system.
  • Negative \( Q \): Heat is released from the system.
In the given exercise, the gas releases 20 J of heat, thus \( Q = -20 \) J. This release results in a decrease in the internal energy of the gas. This process is typical: when a system releases heat, it typically loses energy to the surroundings. In nature, energy spontaneously moves towards spreading out or dispersing unless acted upon by an external work.
work done
Work done is another crucial concept within the realm of thermodynamics. It is defined as the energy transferred when a force is applied over a distance. However, in thermodynamics, it relates more to changes in volume or pressure within a system.

Within the context of the first law of thermodynamics:
  • Work done on the system is considered positive, as it increases the system's energy.
  • Work done by the system is negative, as it leads to energy leaving the system.
In our exercise, 8 J of work is done on the gas. This means an external force is applied, compressing the gas and increasing its internal energy. The work done affects the overall change in internal energy, as shown in the first law's equation \( \Delta U = Q + W \). Here, \( W = 8 \) J is added to the system, working against the energy lost through heat release.

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Most popular questions from this chapter

In a container of negligible heat capacity, \(200 \mathrm{~g}\) ice at \(0^{\circ} \mathrm{C}\) and \(100 \mathrm{~g}\) steam at \(100^{\circ} \mathrm{C}\) are added to \(200 \mathrm{~g}\) of water that has temperature \(55^{\circ} \mathrm{C}\). Assume no heat is lost to the surroundings and the pressure in the container is constant \(1 \mathrm{~atm} .\) Amount of the Sm left in the system, is equal to (A) \(16.7 \mathrm{~g}\) (B) \(8.4 \mathrm{~g}\) (C) \(12 \mathrm{~g}\) (D) \(0 \mathrm{~g}\) Copyright () StemEZ.com. All rights reserved.

Instructions:Read the assertion and reason carefully to mask the correct option out of the options given below. (A) If both assertion and reason are true and the reason is the correct explanation of the assertion. (B) If both assertion and reason are true but reason is not be correct explanation of assertion. (C) If assertion is true but reason is false. (D) If the assertion and reason both are false. Assertion: The carnot is useful in understanding the performance of heat engine Reason: The carnot cycle provides a way of determining the maximum possible efficiency achievable with reservoirs of given temperatures. (A) \(\mathrm{A}\) (B) B (C) \(\mathrm{C}\) (D) \(\mathrm{D}\)

For an adiabatic expansion of a perfect gas, the value of \(\\{(\Delta \mathrm{P}) / \mathrm{P}\\}\) is equal to (A) \(-\sqrt{\gamma}\\{(\Delta \mathrm{r}) / \mathrm{v}\\}\) (B) \(-\\{(\Delta \mathrm{v}) / \mathrm{v}\\}\) (C) \(-\gamma^{2}\\{(\Delta \mathrm{v}) / \mathrm{v}\\}\) (D) \(-\gamma\\{(\Delta \mathrm{v}) / \mathrm{v}\\}\)

The internal energy change in a system that has absorbed 2 Kcal of heat and done \(500 \mathrm{~J}\) of work is (A) \(7900 \mathrm{~J}\) (B) \(4400 \mathrm{~J}\) (C) \(6400 \mathrm{~J}\) (D) \(8900 \mathrm{~J}\)

Carnot engine working between \(300 \mathrm{~K}\) and \(600 \mathrm{~K}\) has work output of \(800 \mathrm{~J}\) per cycle. What is amount of heat energy supplied to the engine from source per cycle (A) \(1600\\{\mathrm{~J} /\) (cycle)\\} (B) \(2000\\{\mathrm{~J} /(\mathrm{cycle})\\}\) (C) \(1000\\{\mathrm{~J} /(\) cycle \()\\}\) (D) \(1800\\{\mathrm{~J} /(\) cycle \()\\}\)

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