Question
Answers

The adiabatic elasticity of a diatomic gas at NTP is:
A.Zero
B.$1 \times {10^5}$${\text{N / }}{{\text{m}}^{\text{2}}}$
C.$1.4 \times {10^5}$${\text{N / }}{{\text{m}}^{\text{2}}}$
D.$2.75 \times {10^5}$${\text{N / }}{{\text{m}}^{\text{2}}}$

Answer
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Hint:The adiabatic elasticity is about the gas which is compressed in such a way that the heat is neither allowed to enter nor it is allowed to leave the system. The formula for adiabatic elasticity is ${{k\Phi }}$.

Complete answer:
Due to the high compressibility of the gases they possess volume elasticity as they can be compressed and expanded very easily. However the magnitude of the volume elasticity depends on the conditions under which it is compressed. When the gas is compressed under conditions under which the temperature is not allowed to change then it is called isothermal elasticity and it is represented by the formula \[{\text{kT}}\].
When the gas is compressed under conditions when no heat is allowed to enter or leave the system then it is called adiabatic elasticity. The formula for adiabatic elasticity is ${{k\Phi }}$. The mathematical formula for the adiabatic elasticity of a gas is equal to ${{{K}}_{{\Phi }}}{{ = \gamma P}}$, where ${{\gamma }}$ is the ratio of the specific heat of gas at constant pressure to the specific heat of a gas at constant volume and P is the pressure of the gas.
The adiabatic elasticity of a diatomic gas at NTP is equal to $1.4 \times {10^5}$.

So, the correct option is C.
Note:
The isothermal elasticity of an ideal gas is equal to ‘\[{{\text{K}}_{\text{T}}}{\text{ = p}}\]’ while the adiabatic elasticity of a gas at Normal conditions of temperature and pressure is equal to ‘${{\text{K}}_{{\Phi }}}{{ = \gamma P}}$’. Hence the ratio of the adiabatic elasticity of a gas to the isothermal elasticity of an ideal gas is equal to ${{\gamma }}$ which the ratio of the specific heat of gas at constant pressure to the specific heat of a gas at constant volume.