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Chapter 15: Problem 10

A system gains 2780 \(\mathrm{J}\) of heat at a constant pressure of \(1.26 \times 10^{5} \mathrm{Pa}\) , and its internal energy increases by 3990 \(\mathrm{J}\) . What is the change in the volume of the system, and is it an increase or a decrease?

Short Answer

Expert verified
Volume decreases by 0.009603 m鲁.

Step by step solution

01

Understand the First Law of Thermodynamics

The first law of thermodynamics states that the change in internal energy \((\Delta U)\) of a system is equal to the heat added to the system \((Q)\) minus the work done by the system \((W)\). Mathematically, it is expressed as:\[ \Delta U = Q - W \]
02

Calculate the Work Done by the System

Rearranging the formula from Step 1, we can solve for work done:\[ W = Q - \Delta U \]Here, given \( Q = 2780 \ \mathrm{J} \) and \( \Delta U = 3990 \ \mathrm{J} \), substitute these values:\[ W = 2780 \ \mathrm{J} - 3990 \ \mathrm{J} = -1210 \ \mathrm{J} \]The negative sign indicates work was done on the system.
03

Calculate Change in Volume

The work done by a system at constant pressure \(P\) is given by:\[ W = P \Delta V \]where \( \Delta V \) is the change in volume. Rearrange this equation to solve for \( \Delta V \):\[ \Delta V = \frac{W}{P} \]Substitute the known values \( W = -1210 \ \mathrm{J} \) and \( P = 1.26 \times 10^5 \ \mathrm{Pa} \):\[ \Delta V = \frac{-1210 \ \mathrm{J}}{1.26 \times 10^5 \ \mathrm{Pa}} \]\[ \Delta V = -0.009603 \ \mathrm{m^3} \]
04

Interpret the Sign of Change in Volume

The negative sign of \( \Delta V \) indicates a decrease in volume. Thus, the volume of the system has decreased by 0.009603 \( \mathrm{m^3} \).

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Key Concepts

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

Internal Energy
Internal energy is a key concept in thermodynamics. It refers to the total energy stored within a system. This includes energies related to the random motion of molecules, chemical bonds, and other internal forces. Internal energy changes can occur due to:
  • Addition or removal of heat
  • Work done by or on the system
In our example, we see an increase in internal energy by 3990 J. This happens because the system is gaining energy from heat, increasing its total internal energy. Internal energy is a state function, meaning it depends only on the current state of the system, not on how that state was reached.
Heat Transfer
Heat transfer is the process of energy being transferred due to temperature differences. In the discussed scenario, 2780 J of heat is added to the system. Heat transfer can occur in several ways:
  • Conduction: Direct contact transfer
  • Convection: Heat transfer through fluids
  • Radiation: Energy transfer through electromagnetic waves
This transfer of heat is one part of the First Law of Thermodynamics equation, which accounts for the change in internal energy and work done.
Work Done
Work done by a system can alter its internal energy. It happens when a force causes displacement in the system's volume. In our example, the calculation of work done is essential:
  • Work is expressed as: \( W = Q - \Delta U \)
  • Here, \( W = 2780 \, \text{J} - 3990 \, \text{J} = -1210 \, \text{J} \)
The negative work indicates that the surroundings did work on the system. Work done on the system increases its internal energy, which is why we observe an increase despite losing some amount through work.
Volume Change
Volume change in a system can be a result of work being done. In thermodynamics, this is observed under constant pressure conditions. For our scenario, the volume change \( \Delta V \) is calculated using the equation:
\[ \Delta V = \frac{W}{P} \] Where \( P \) is constant pressure. Substituting in the values:
\[ \Delta V = \frac{-1210 \, \text{J}}{1.26 \times 10^5 \, \text{Pa}} = -0.009603 \, \text{m}^3 \] This negative result shows the system's volume decreases as work is done on it. Hence, the space occupied by the system is reduced.

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Most popular questions from this chapter

A system gains 1500 \(\mathrm{J}\) of heat, while the internal energy of the system increases by 4500 \(\mathrm{J}\) and the volume decreases by 0.010 \(\mathrm{m}^{3}\) . Assume that the pressure is constant and find its value.

One-half mole of a monatomic ideal gas expands adiabatically and does 610 \(\mathrm{J}\) of work. By how many kelvins does its temperature change? Specify whether the change is an increase or a decrease.

(a) After 6.00 \(\mathrm{kg}\) of water at \(85.0^{\circ} \mathrm{C}\) is mixed in a perfect thermos with 3.00 \(\mathrm{kg}\) of ice at \(0.0^{\circ} \mathrm{C},\) the mixture is allowed to reach equilibrium. When heat is added to or removed from a solid or liquid of mass \(m\) and specific heat capacity \(c\) , the change in entropy can be shown to be \(\Delta S=m c \ln \left(T_{\mathrm{f}} / T_{\mathrm{i}}\right),\) where \(T_{\mathrm{i}}\) and \(T_{\mathrm{f}}\) are the initial and final Kelvin temperatures. Using this expression and the change in entropy for melting, find the change in entropy that occurs. \((\mathrm{b})\) Should the entropy of the universe increase or decrease as a result of the mixing process? Give your reasoning and state whether your answer in part (a) is consistent with your answer here.

The sun is a sphere with a radius of \(6.96 \times 10^{8} \mathrm{m}\) and an average surface temperature of 5800 k. Determine the amount by which the sun's thermal radiation increases the entropy of the entire universe each second. Assume that the sun is a perfect blackbody, and that the average temperature of the rest of the universe is 2.73 \(\mathrm{K}\) . Do not consider the thermal radiation absorbed by the sun from the rest of the universe.

The drawing (not to scale) shows the way in which the pressure and volume change for an ideal gas that is used as the working substance in a Carnot engine. The gas begins at point a (pressure \(=P_{a},\) volume \(=V_{\mathrm{a}}\) and expands isothermally at temperature \(T_{\mathrm{H}}\) until point b (pressure \(=P_{\mathrm{b}}\) , volume \(=V_{\mathrm{b}}\) is reached. During this expansion, the input heat of magnitude \(\left|Q_{\mathrm{H}}\right|\) enters the gas from the hot reservoir of the engine. Then, from point b to point \(c\) (pressure \(=P_{c},\) volume \(=V_{c} )\) the gas expands adiabatically. Next, the gas is compressed isothermally at temperature \(T_{\mathrm{C}}\) from point \(c\) to point \(\mathrm{d}\) (pressure \(=P_{\mathrm{d}},\) volume \(=V_{\mathrm{d}} )\) During this compression, heat of magnitude \(\left|Q_{\mathrm{C}}\right|\) is rejected to the cold reservoir of the engine. Finally, the gas is compressed adiabatically from point d to point a, where the gas is back in its initial state. The overall process a to b to c to d to a is called a Carnot cycle. Prove for this cycle that \(\left|Q_{\mathrm{c}}\right| /\left|Q_{\mathrm{H}}\right|=T_{\mathrm{C}} / T_{\mathrm{H}} .\)
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