The ratio of two specific heats of gas $\dfrac{{{C_P}}}{{{C_V}}}$ for argon is 1.6 and for hydrogen is 1.4. If adiabatic elasticity of argon at pressure P is E , at what pressure the adiabatic elasticity of hydrogen will also equal to E?
A) $P$
B) $1.4P$
C) $\dfrac{7}{8}P$
D) $\dfrac{8}{7}P$
Answer
Verified
117.9k+ views
Hint: In an adiabatic process, heat is neither added to the system nor it escapes the system. The pressure at which this happens is called the adiabatic pressure. To obtain adiabatic pressure, a number known as adiabatic constant $\gamma $ should be multiplied.
Complete step by step solution:
The adiabatic elasticity of a gas is given by-
$E = \gamma P$ Where E=elasticity, $\gamma $= ratio of two specific heats,
P= pressure of the gas.
For argon given that ${\gamma _{Ar}} = {\left( {\dfrac{{{C_P}}}{{{C_V}}}} \right)_{Ar}} = 1.6$
Therefore,
$E = {\gamma _{Ar}}P$
$ \Rightarrow E = 1.6P$
Let the pressure required for hydrogen is ${P'}$
Now given that
${\gamma _{{H_2}}} = {\left( {\dfrac{{{C_P}}}{{{C_V}}}} \right)_{_{{H_2}}}} = 1.4$
As elasticity of hydrogen is also E, So
$E = {\gamma _{{H_2}}}{P_0}$
$ \Rightarrow E = 1.4{P_0}$
Now compare the above equations, we get –
$1.4{P_0} = 1.6P$
$ \Rightarrow {P_0} = \dfrac{{1.6}}{{1.4}}P$
$ \Rightarrow {P_0} = \dfrac{8}{7}P$
Hence, the correct option is Option C.
Additional information: An adiabatic process is defined as, The thermodynamic process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression.
The adiabatic process can be either reversible or irreversible.
Following are the essential conditions for the adiabatic process to take place:
1. The system must be perfectly insulated from the surrounding.
2. The process must be carried out quickly so that there is a sufficient amount of time for heat transfer to take place.
Note: The students should always remember that the value of ${C_P} > {C_V}$ and the relation between these two specific heats is given by Mayor’s Formula i.e. ${C_P} - {C_V} = R$ where R= universal gas constant.
Complete step by step solution:
The adiabatic elasticity of a gas is given by-
$E = \gamma P$ Where E=elasticity, $\gamma $= ratio of two specific heats,
P= pressure of the gas.
For argon given that ${\gamma _{Ar}} = {\left( {\dfrac{{{C_P}}}{{{C_V}}}} \right)_{Ar}} = 1.6$
Therefore,
$E = {\gamma _{Ar}}P$
$ \Rightarrow E = 1.6P$
Let the pressure required for hydrogen is ${P'}$
Now given that
${\gamma _{{H_2}}} = {\left( {\dfrac{{{C_P}}}{{{C_V}}}} \right)_{_{{H_2}}}} = 1.4$
As elasticity of hydrogen is also E, So
$E = {\gamma _{{H_2}}}{P_0}$
$ \Rightarrow E = 1.4{P_0}$
Now compare the above equations, we get –
$1.4{P_0} = 1.6P$
$ \Rightarrow {P_0} = \dfrac{{1.6}}{{1.4}}P$
$ \Rightarrow {P_0} = \dfrac{8}{7}P$
Hence, the correct option is Option C.
Additional information: An adiabatic process is defined as, The thermodynamic process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression.
The adiabatic process can be either reversible or irreversible.
Following are the essential conditions for the adiabatic process to take place:
1. The system must be perfectly insulated from the surrounding.
2. The process must be carried out quickly so that there is a sufficient amount of time for heat transfer to take place.
Note: The students should always remember that the value of ${C_P} > {C_V}$ and the relation between these two specific heats is given by Mayor’s Formula i.e. ${C_P} - {C_V} = R$ where R= universal gas constant.
Recently Updated Pages
JEE Main 2025: Application Form, Exam Dates, Eligibility, and More
Uniform Acceleration - Definition, Equation, Examples, and FAQs
How to find Oxidation Number - Important Concepts for JEE
How Electromagnetic Waves are Formed - Important Concepts for JEE
Electrical Resistance - Important Concepts and Tips for JEE
Average Atomic Mass - Important Concepts and Tips for JEE
Trending doubts
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Chemistry Question Paper with Answer Keys and Solutions
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
Physics Average Value and RMS Value JEE Main 2025
Inductive Effect and Acidic Strength - Types, Relation and Applications for JEE
Free Radical Substitution Mechanism of Alkanes for JEE Main 2025
Other Pages
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids
NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion
Thermodynamics Class 11 Notes CBSE Physics Chapter 11 (Free PDF Download)
NCERT Solutions for Class 11 Physics Chapter 5 Work Energy and Power
NCERT Solutions for Class 11 Physics Chapter 3 Motion In A Plane