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Q12.4:
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 ?

Solution :

Step 1: Analyzing the Thermodynamic Process

The problem states that the cylinder walls and the piston are made of heat insulators. This implies that no heat exchange occurs between the system and the surroundings ($dQ = 0$). [Per the definition of an adiabatic process]. Consequently, the compression of the hydrogen gas follows the adiabatic relation between pressure ($P$) and volume ($V$).

Step 2: Defining Physical Constants for Hydrogen

Hydrogen ($H_2$) is a diatomic gas. For a diatomic gas, the degree of freedom ($f$) is 5 at room temperature. The adiabatic exponent ($\gamma$), defined as the ratio of molar heat capacity at constant pressure ($C_p$) to constant volume ($C_v$), is calculated as follows:

$\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f} = 1 + \frac{2}{5} = 1.4$

Step 3: Applying the Adiabatic Gas Law

The state equation for an adiabatic process is given by: $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$ [Per the Adiabatic Equation of State]

Given the condition that the gas is compressed to half its original volume: $V_2 = \frac{1}{2} V_1$

Step 4: Algebraic Derivation of the Pressure Factor

Rearranging the adiabatic equation to find the ratio of final pressure ($P_2$) to initial pressure ($P_1$): $\frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^{\gamma}$

Substituting $V_2 = \frac{V_1}{2}$ into the equation: $\frac{P_2}{P_1} = \left( \frac{V_1}{V_1 / 2} \right)^{1.4}$

$\frac{P_2}{P_1} = (2)^{1.4}$

Step 5: Numerical Calculation

To evaluate $(2)^{1.4}$:

$(2)^{1.4} \approx 2.639$

V P Adiabatic Curve: PV^{1.4} = constant State 1 (V_1, P_1) State 2 (V_1/2, P_2)

Summary Table of Parameters

Parameter Value
Adiabatic Exponent ($\gamma$) $1.4$
Volume Compression Ratio $2$
Pressure Factor $2^{1.4} \approx 2.64$

Final Solution: The pressure of the gas increases by a factor of approximately 2.64.


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