Answer
64.8k+ views
Hint: Since this is a choice-based question, we will check each option one by one and eliminate accordingly. We will use the ideal gas equation, \[{\text{PV = nRT}}\] and then check the relation of specific heats at constant pressure and volume one by one.
Complete step by step solution:
Let’s begin by checking the validity of option a.) A gas has two specific heats only.
We know that,
\[{\text{PV = nRT}}\] (Where, P is pressure; V is the volume; n is the amount of substance; R is the ideal gas constant)
Now, at constant pressure, we define specific heat in the form of \[{{\text{C}}_{\text{p}}}\] and at constant volume, specific heat is defined as \[{{\text{C}}_{\text{v}}}\] .
But, at constant temperature, there is no change in the temperature which means there exists an isothermal condition. And since the temperature is constant, there won’t be any change in the specific heat either. Therefore, we can say, this process does not involve any specific heat.
Hence, a gas has only two specific heats, one at constant temperature and other at constant pressure. So, option (a) is correct.
Now, let’s check option (b).
As stated in the question, the coefficient of thermal expansion ${\alpha _t}$
From the expression, $\dfrac{{\Delta L}}{L} = {\alpha _t}\Delta T$ we got to know that, the coefficient of thermal expansion measures the deforming capability of a material.
This proves that the coefficient of thermal expansion has no relation with the specific heats. It is an independent quantity.
Since we have proved option (a) correct, therefore option (c) and (d) are incorrect.
Note: Specific heat also depends on the conditions of the experiment, i.e., the way in which heat is supplied to the body. In general, experiments are made either at constant volume or at constant pressure.
In case of solids and liquids, due to small thermal expansion, the difference in measured values of specific heat is very small and is usually neglected. However, in case of gases, specific heat at constant volume is quite different from that at constant pressure.
Complete step by step solution:
Let’s begin by checking the validity of option a.) A gas has two specific heats only.
We know that,
\[{\text{PV = nRT}}\] (Where, P is pressure; V is the volume; n is the amount of substance; R is the ideal gas constant)
Now, at constant pressure, we define specific heat in the form of \[{{\text{C}}_{\text{p}}}\] and at constant volume, specific heat is defined as \[{{\text{C}}_{\text{v}}}\] .
But, at constant temperature, there is no change in the temperature which means there exists an isothermal condition. And since the temperature is constant, there won’t be any change in the specific heat either. Therefore, we can say, this process does not involve any specific heat.
Hence, a gas has only two specific heats, one at constant temperature and other at constant pressure. So, option (a) is correct.
Now, let’s check option (b).
As stated in the question, the coefficient of thermal expansion ${\alpha _t}$
From the expression, $\dfrac{{\Delta L}}{L} = {\alpha _t}\Delta T$ we got to know that, the coefficient of thermal expansion measures the deforming capability of a material.
This proves that the coefficient of thermal expansion has no relation with the specific heats. It is an independent quantity.
Since we have proved option (a) correct, therefore option (c) and (d) are incorrect.
Note: Specific heat also depends on the conditions of the experiment, i.e., the way in which heat is supplied to the body. In general, experiments are made either at constant volume or at constant pressure.
In case of solids and liquids, due to small thermal expansion, the difference in measured values of specific heat is very small and is usually neglected. However, in case of gases, specific heat at constant volume is quite different from that at constant pressure.
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