Question

Thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If $V_2=2V_1$ then the ratio of temperature $T_2/T_1$ is: The process is $PV^{1/2}$ = constant. 

 

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $\frac{1}{2}$
• D. 2

Hint:

For any polytropic process of the form $PV^n = \text{constant}$ for an ideal gas, a useful derived relation is $TV^{n-1} = \text{constant}$. In this problem, $n=1/2$, so we get $TV^{1/2-1} = TV^{-1/2} = \text{constant}$, which quickly leads to the solution.

Answer 1

The Correct Option is B

The given thermodynamic process is a polytropic process described by $PV^{1/2} = k$ (constant).
For one mole of an ideal gas, the ideal gas equation is $PV = RT$.
We can express the pressure $P$ from the ideal gas equation as $P = \frac{RT}{V}$.
Substitute this expression for $P$ into the process equation:
$\left(\frac{RT}{V}\right)V^{1/2} = k$.
$RTV^{-1}V^{1/2} = k \implies RTV^{-1/2} = k$.
Since R is a constant, we have the relation $TV^{-1/2} = \text{constant}$.
This means that for the initial state (1) and final state (2), we have:
$T_1V_1^{-1/2} = T_2V_2^{-1/2}$.
We need to find the ratio $T_2/T_1$. Rearranging the equation:
$\frac{T_2}{T_1} = \frac{V_1^{-1/2}}{V_2^{-1/2}} = \left(\frac{V_2}{V_1}\right)^{1/2}$.
Given that $V_2 = 2V_1$, the ratio $\frac{V_2}{V_1} = 2$.
Substituting this into our expression for the temperature ratio:
$\frac{T_2}{T_1} = (2)^{1/2} = \sqrt{2}$.