Question

An ideal gas is initially at a temperature T and volume V. its volume is increased by $\Delta V$ by increasing the temperature by $\Delta T$ then the factor $\delta = \dfrac{{\Delta V}}{{V\Delta T}}$ varies with temperature as:
A) $\delta$ $\alpha$ $\dfrac{1}{{{T^2}}}$
B) $\delta$ $\alpha$ $T$
C) $\delta$ $\alpha$ $\dfrac{1}{T}$
D) $\delta$ $\alpha$ $\sqrt T$

Answer 1

The application of the first law of thermodynamics plays a very important role in this question. The Gas law is also very important for this question.

Complete step by step solution:
The first law of thermodynamics states that the change in heat of a system is equal to the sum of change in its internal energy and the thermodynamic work done during the process.
And the gas law states that the product of pressure and volume at a specific temperature is directly proportional to that temperature.
Now, according to the first law of thermodynamics;
$\Delta Q = \Delta U + W$
But at constant pressure the change in heat is equal to the change in entropy.
So, $\Delta Q = \Delta H$
At constant pressure the first law can be modified as;
$\Rightarrow \Delta H = \Delta U + W$ (Equation: 1)
Now we all know that;
The thermodynamic work done is mathematically written as;
$W = P\Delta V$
Now since the volume is increasing the work done will be positive. This is because the process is an expansion process.
$\Rightarrow \Delta H = \Delta U + P\Delta V$(Equation: 2)
Now we know that;
$\Delta H = n{C_p}\Delta T$
And $\Delta U = n{C_v}\Delta T$
$\Rightarrow n{C_p}\Delta T = n{C_v}\Delta T + P\Delta V$ (Equation: 3)
$\Rightarrow P\Delta V = n({C_p} - {C_v})\Delta T$
Now as ${C_p} - {C_v} = R$ (Equation: 4)
Thus, from equation 3 and equation 4 we get;
$\Rightarrow P\Delta V = nR\Delta T$
$\Rightarrow \dfrac{{\Delta V}}{{\Delta T}} = \dfrac{{nR}}{P}$ (Equation: 5)
And according to the gas law pressure at any temperature is given by;
$P = \dfrac{{nRT}}{V}$ (Equation: 6)
So, from equation 5 and equation 6 we get;
$\Rightarrow \dfrac{{\Delta V}}{{V\Delta T}} = \dfrac{1}{T}$
$\therefore \delta = \dfrac{1}{T}$

Therefore, option C is correct.

Note: Work done in the expansion process is positive, the final volume is greater than initial volume.
Work done in the compression process is negative as the final volume is less than the initial volume.
Work done on the system is negative.
Work done by the system is positive.