Question

A cylinder with a movable piston contains $3$ moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Answer 1

Hint: A gas can do work, expansion or compression, through various thermodynamic processes. In this, we have hydrogen gas that undergoes an adiabatic process; therefore, we will apply the thermodynamic equation for an adiabatic process relating the initial and final, and pressure and volume of the gas. The ratio of specific heats for diatomic is considered as $\gamma =1.4$.

Formula used:
For an Adiabatic process,
${{P}_{1}}{{\left( {{V}_{1}} \right)}^{\gamma }}={{P}_{2}}{{\left( {{V}_{2}} \right)}^{\gamma }}$
For diatomic gas,
$\gamma =1.4$

Complete step by step answer:
Thermodynamic process is the passage of a thermodynamic system from an initial state to a final state of thermodynamic equilibrium. The initial and the final states of a system are the defining elements of a thermodynamic process. The four types of thermodynamic processes are: Isothermal, Isochoric, Isobaric and Adiabatic.
In an isothermal process, the temperature of a system remains constant. Thermal equilibrium is maintained during the process. The equation of the Isothermal process is given by,
$PV=\text{ constant}$
In an isochoric process, the volume of the closed system undergoing the process remains unchanged. The Isochoric process is an Isovolumetric process. The equation of the Isochoric process is given by,
$\dfrac{P}{T}=\text{ constant}$
In an Isobaric process, the pressure of the system stays constant. The equation of the Isobaric process is given by,
 $\dfrac{V}{T}=\text{ constant}$
In an adiabatic process, the heat and mass transfer between system and surroundings remains zero. The equation of the adiabatic process is given by,
$P{{V}^{\gamma }}=\text{ constant}$
Where,
$P$ is the pressure of the system
$V$ is the volume of the system
$T$ is the temperature of the system
$\gamma $is the ratio of specific heats $\left( \gamma =\dfrac{{{C}_{P}}}{{{C}_{V}}} \right)$
As we are given that the cylinder is completely insulated from its surroundings; therefore, no heat will be exchanged between the system (Cylinder) and its surroundings. Thus, the given process is adiabatic.
Initial pressure inside the cylinder is ${{P}_{1}}$
Final pressure inside the cylinder is ${{P}_{2}}$
Initial volume inside the cylinder is ${{V}_{1}}$
Final volume inside the cylinder is ${{V}_{2}}$
Ratio of specific heats is given as,
$\gamma =\dfrac{{{C}_{p}}}{{{C}_{v}}}$
For diatomic gas, $\gamma =1.4$
Now,
Expression for an adiabatic process is given as,
${{P}_{1}}{{\left( {{V}_{1}} \right)}^{\gamma }}={{P}_{2}}{{\left( {{V}_{2}} \right)}^{\gamma }}$
We have,
${{V}_{2}}=\dfrac{{{V}_{1}}}{2}$
Therefore,
$\begin{align}
  & {{P}_{1}}{{\left( {{V}_{1}} \right)}^{\gamma }}={{P}_{2}}{{\left( \dfrac{{{V}_{1}}}{2} \right)}^{\gamma }} \\
 & \dfrac{{{P}_{2}}}{{{P}_{1}}}=\dfrac{{{\left( {{V}_{1}} \right)}^{\gamma }}}{{{\left( \dfrac{{{V}_{1}}}{2} \right)}^{\gamma }}} \\
 & \dfrac{{{P}_{2}}}{{{P}_{1}}}={{2}^{\gamma }} \\
\end{align}$
Put $\gamma =1.4$
$\begin{align}
  & \dfrac{{{P}_{2}}}{{{P}_{1}}}={{\left( 2 \right)}^{1.4}}=2.639 \\
 & {{P}_{2}}=2.639{{P}_{1}} \\
\end{align}$
Hence, the pressure increases by a factor of $2.639$.

Note:
Students need to remember the concept of the thermodynamic processes and the respective expressions for them to solve the above equation. Also, keep in mind that in an Adiabatic process, heat and mass transfer between system and surroundings remain zero. The equation for the relation between initial and final; pressure and volume can be applied accordingly.
The ratio of specific heats for different types of gases:
For monoatomic gas, $\gamma =1.66$
For diatomic gas, $\gamma =1.4$
For triatomic gas, $\gamma =1.28$