If 0.1kcal of heat is added to 1.2L of ${O_2}$ in a cylinder at constant pressure of 1 atm, the volume increases to 1.5L. Calculate $\Delta U$ and $\Delta H$ of the process (1L-atm = 100J).
Answer
593.7k+ views
Hint: When heat is added to the system at constant pressure, the change in enthalpy is equal to the heat flow. The internal energy increases when heat is provided to the system.
Formula used: $\Delta U = q - P\Delta V$ and $\Delta H = {q_p}$
Complete step-by-step solution:
As we know, change in internal energy of a system is the sum of heat and work. The quantity of heat transferred from the surroundings to the system is q and the work done in the process is w, then the change in internal energy is given by
$\Delta U = Q + W$
In the above equation, work done can be represented as expansion and compression of gases, written as
$W = P\Delta V$
Looking at the change in volume, we can determine whether it is expansion or compression.
In a closed container, if a piston is being pushed down by supplying energy, compression of the gas occurs resulting in a decrease in volume. This tells us that work is done on the system and it is positive. And if the volume increases on expansion of gas against an external pressure and the piston comes up, that means work is done by the system and it is negative.
Here, as the volume is increasing from 1.2L to 1.5L, we say, work is done by the system.
Therefore, the formula of change in internal energy now becomes
$\Delta U = q - P\Delta V$
Given: heat (q)=1 kcal i.e. 1000cal
Initial volume \[({V_1}) = 1.2L\]
Final volume \[({V_2}) = 1.5L\]
pressure (P) =1 atm
Putting the given values in the above formula, we get
$\Delta U = 1000cal - [1atm(1.5 - 1.2)L]$
$\Rightarrow$ $\Delta U = 992.7cal \, or \, 0.993kcal$
Now, we have to calculate $\Delta H$
At constant pressure,
We know that, $\Delta H = {q_p}$
We know that, $\Delta H = {q_p}$
Therefore, \[\Delta H = 1kcal\]
Hence, for the given system, $\Delta U = 992.7cal \,or \, 0.993\,kcal$ and \[\Delta H = 1kcal\]
Note: The change in internal energy states the first law of thermodynamics. According to this law, energy can neither be created nor be destroyed; it can only be transformed from one form to another. Applications of this law include adiabatic, isochoric, isothermal, isobaric and free expansion processes.
Formula used: $\Delta U = q - P\Delta V$ and $\Delta H = {q_p}$
Complete step-by-step solution:
As we know, change in internal energy of a system is the sum of heat and work. The quantity of heat transferred from the surroundings to the system is q and the work done in the process is w, then the change in internal energy is given by
$\Delta U = Q + W$
In the above equation, work done can be represented as expansion and compression of gases, written as
$W = P\Delta V$
Looking at the change in volume, we can determine whether it is expansion or compression.
In a closed container, if a piston is being pushed down by supplying energy, compression of the gas occurs resulting in a decrease in volume. This tells us that work is done on the system and it is positive. And if the volume increases on expansion of gas against an external pressure and the piston comes up, that means work is done by the system and it is negative.
Here, as the volume is increasing from 1.2L to 1.5L, we say, work is done by the system.
Therefore, the formula of change in internal energy now becomes
$\Delta U = q - P\Delta V$
Given: heat (q)=1 kcal i.e. 1000cal
Initial volume \[({V_1}) = 1.2L\]
Final volume \[({V_2}) = 1.5L\]
pressure (P) =1 atm
Putting the given values in the above formula, we get
$\Delta U = 1000cal - [1atm(1.5 - 1.2)L]$
$\Rightarrow$ $\Delta U = 992.7cal \, or \, 0.993kcal$
Now, we have to calculate $\Delta H$
At constant pressure,
We know that, $\Delta H = {q_p}$
We know that, $\Delta H = {q_p}$
Therefore, \[\Delta H = 1kcal\]
Hence, for the given system, $\Delta U = 992.7cal \,or \, 0.993\,kcal$ and \[\Delta H = 1kcal\]
Note: The change in internal energy states the first law of thermodynamics. According to this law, energy can neither be created nor be destroyed; it can only be transformed from one form to another. Applications of this law include adiabatic, isochoric, isothermal, isobaric and free expansion processes.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

In what year Guru Nanak Dev ji was born A15 April 1469 class 11 social science CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

