
An ideal gas has an initial pressure of $3$ pressure units and an initial volume of $4$ volume units. The table gives the final pressure and volume of the gas (in those same units) in four processes. Which processes start and end on the same isotherm:
A B C D Pressure \[5\] $4$ $12$ $6$ Volume $7$ $6$ $1$ $3$
a. A
b. B
c. .C
d. D
A | B | C | D | |
Pressure | \[5\] | $4$ | $12$ | $6$ |
Volume | $7$ | $6$ | $1$ | $3$ |
Answer
474.3k+ views
Hint:We can try to approach the solution to this problem by thinking about the process’ curves and the condition for the isothermal process. An isothermal process means that the value of temperature should remain constant during the whole process, which in turn means that the product of values of pressure and volume remains constant.
Complete step-by-step solution:
We will try to solve the question just like we explained in the hint. We will first give thought to what conditions every process follows. Then we will compare the given values in the question to reach at the correct answer or option.
The question asks about an ideal gas on an isothermal curve. So, we need to think about the condition that an isothermal process follows. An isothermal process requires the temperature to remain constant during the whole process.
We already know through the ideal gas equation:
$PV\, = \,nRT$
Where, $P$ stands for pressure
$V$ stands for volume
$R$ is ideal gas constant
$T$ stands for temperature
$n$ represents number of moles
We know that $R$ is a universal constant, $n$ is constant in the process as the gas is not inserted or removed.
For an isothermal process, $T$ needs to be a constant too.
$PV\, = \,nRT\, = \,k$
Where, $k$ is a constant value
Hence, the product of values of pressure and volume needs to remain constant as well.
Let us find out the product value in initial state:
Given in the question, for the initial state:
We have been given the value of pressure, $P\, = \,3$
The value of volume given to us is, $V\, = \,4$
Substituting the values, we get the value of product as:
${k_i}\, = \,12\,units$
Now, we will find out the value of the product for each case and compare with its initial value.
For case A:
We have been given the value of pressure, $P\, = \,5$
The value of volume given to us is, $V\, = \,7$
Substituting the values, we get the value of product as:
${k_A}\, = \,35\,units$
We can clearly see that this is not equal to the initial value, hence, Case A is the wrong option.
For case B:
We have been given the value of pressure, $P\, = \,4$
The value of volume given to us is, $V\, = \,6$
Substituting the values, we get the value of product as:
${k_B}\, = \,24\,units$
We can see that this value is not the same as the initial value, hence case B is also the wrong option.
For case C:
We have been given the value of pressure, $P\, = \,12$
The value of volume given to us is, $V\, = \,1$
Substituting the values, we get the value of product as:
${k_C}\, = \,12\,units$
The value for the product of pressure and volume is the same as the initial value, hence this is the correct option.
The correct answer is option C.
Note:- Many students try to approach the solution to this question with more complex and non-basic methods, which only makes it tougher to reach the answer. Also, try to remember the conditions for other processes as well, like adiabatic processes whose condition is that the value of heat exchange value is zero.
Complete step-by-step solution:
We will try to solve the question just like we explained in the hint. We will first give thought to what conditions every process follows. Then we will compare the given values in the question to reach at the correct answer or option.
The question asks about an ideal gas on an isothermal curve. So, we need to think about the condition that an isothermal process follows. An isothermal process requires the temperature to remain constant during the whole process.
We already know through the ideal gas equation:
$PV\, = \,nRT$
Where, $P$ stands for pressure
$V$ stands for volume
$R$ is ideal gas constant
$T$ stands for temperature
$n$ represents number of moles
We know that $R$ is a universal constant, $n$ is constant in the process as the gas is not inserted or removed.
For an isothermal process, $T$ needs to be a constant too.
$PV\, = \,nRT\, = \,k$
Where, $k$ is a constant value
Hence, the product of values of pressure and volume needs to remain constant as well.
Let us find out the product value in initial state:
Given in the question, for the initial state:
We have been given the value of pressure, $P\, = \,3$
The value of volume given to us is, $V\, = \,4$
Substituting the values, we get the value of product as:
${k_i}\, = \,12\,units$
Now, we will find out the value of the product for each case and compare with its initial value.
For case A:
We have been given the value of pressure, $P\, = \,5$
The value of volume given to us is, $V\, = \,7$
Substituting the values, we get the value of product as:
${k_A}\, = \,35\,units$
We can clearly see that this is not equal to the initial value, hence, Case A is the wrong option.
For case B:
We have been given the value of pressure, $P\, = \,4$
The value of volume given to us is, $V\, = \,6$
Substituting the values, we get the value of product as:
${k_B}\, = \,24\,units$
We can see that this value is not the same as the initial value, hence case B is also the wrong option.
For case C:
We have been given the value of pressure, $P\, = \,12$
The value of volume given to us is, $V\, = \,1$
Substituting the values, we get the value of product as:
${k_C}\, = \,12\,units$
The value for the product of pressure and volume is the same as the initial value, hence this is the correct option.
The correct answer is option C.
Note:- Many students try to approach the solution to this question with more complex and non-basic methods, which only makes it tougher to reach the answer. Also, try to remember the conditions for other processes as well, like adiabatic processes whose condition is that the value of heat exchange value is zero.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Master Class 9 Science: Engaging Questions & Answers for Success

Master Class 9 Social Science: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
Types of lever in which effort is in between fulcrum class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

A two input XOR Gate produces a high output only when class 12 physics CBSE

What is a transformer Explain the principle construction class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE
