Answer to Question #101583 in Molecular Physics | Thermodynamics for Sridhar

Question

An ideal gas at initial temperature (T) and initial volume(V) is expanded adiabatically to a volume (2V). The gas is then expanded isothermally a volume (5V) and thereafter compressed adiabatically so that the temperature of the gas becomes again (T).If the final volume of the gas is x(V). then the value of x is......

Answer 1

The combined gas law:

$\frac{p \cdot V}{T} = const$,

where $p$ is the pressure of the gas, $V$ is the volume of the gas, $T$ is the temperature of the gas.

Initial parameters: $p_1=P$, $V_1=V$ and $T_1=T$.

1) adiabatic process is characterized by the constant entropy.

Gas can be characterized by the parameters $p_2$, $V_2$ and $T_2$.

$V_2=2V$

$T_2=(\frac{V_2}{V_1})^{1 - \gamma} \cdot T_1 = (\frac{2V}{V})^{1 - \gamma} \cdot T= (2)^{1 - \gamma} \cdot T$

where $\gamma$ is the heat capacity ratio, which is constant for a calorifically perfect gas.

2) isothermal process is characterized by the constant temperature.

Gas can be characterized by the parameters $p_3$, $V_3$ and $T_3$.

$V_3=5V$

$T_3=T_2= (2)^{1 - \gamma} \cdot T$

3) adiabatic process is characterized by the constant entropy.

Gas can be characterized by the parameters $p_4$, $V_4$ and $T_4$.

$T_4=T$, so $\frac{T_4}{T_3}=\frac{T}{(2)^{1 - \gamma} \cdot T} = (2)^{ \gamma - 1}$

$V_4=V_3 (\frac{T_4}{T_3})^{\frac{1}{1- \gamma}} = 5V ((2)^{ \gamma - 1})^{\frac{1}{1- \gamma}}=(2)^{\frac{\gamma -1}{1 - \gamma}} \cdot 5V= (2)^{-1} \cdot 5V = \frac{5V}{2} = 2.5V$

Answer: If the final volume of the gas is $x(V)$, then the value of $x$ is 2.5.