A monoatomic gas having $\gamma = \frac{5}{3}$ is stored in a thermally insulated container and the gas is suddenly compressed to $(\frac{1}{8})^{\text{th}}$ of its initial volume. The ratio of final pressure and initial pressure is:
Options: A) 16, B) 32, C) 28, D) 40
Options: A) 16, B) 32, C) 28, D) 40
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Solution Steps
1
Identify the Process Type
The container is “thermally insulated” and the compression is “sudden”. Both these conditions indicate an adiabatic process where no heat exchange occurs with the surroundings ($dQ = 0$).
2
Recall Adiabatic Relation
For an adiabatic process, the relationship between pressure ($P$) and volume ($V$) is given by Poisson’s Law:
$$P V^{\gamma} = \text{constant}$$
Therefore, $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.
3
List Given Values
Initial Volume = $V_1$
Final Volume $V_2 = \frac{1}{8} V_1$
Adiabatic exponent $\gamma = \frac{5}{3}$ (for monoatomic gas).
4
Set up the Ratio Equation
We need to find the ratio $\frac{P_2}{P_1}$. Rearranging the formula:
$$\frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^{\gamma}$$
5
Substitute and Calculate
Substitute $V_2 = \frac{V_1}{8}$ into the equation:
$$\frac{P_2}{P_1} = \left( \frac{V_1}{V_1/8} \right)^{5/3} = (8)^{5/3}$$
6
Final Computation
Express 8 as a power of 2: $8 = 2^3$.
$$\frac{P_2}{P_1} = (2^3)^{5/3} = 2^{3 \times \frac{5}{3}} = 2^5 = 32$$
Final Answer: 32
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Theory
1. Adiabatic Processes
An adiabatic process is a thermodynamic transformation in which no heat is exchanged between the system and its surroundings ($Q=0$). This typically occurs in systems that are either perfectly insulated or when the process happens so rapidly that there is no time for heat transfer to take place. In the context of gases, work done on the gas during adiabatic compression results in an increase in internal energy, leading to a rise in temperature. The governing equation is $PV^\gamma = K$, where $\gamma$ is the ratio of specific heats ($C_p/C_v$). This relation is derived from the first law of thermodynamics and the ideal gas law under the constraint that $dQ=0$.
2. Monoatomic Gas Properties
A monoatomic gas consists of single atoms rather than molecules. Common examples include noble gases like Helium or Argon. According to the law of equipartition of energy, a monoatomic gas has 3 degrees of freedom (all translational). Consequently, its molar specific heat at constant volume ($C_v$) is $\frac{3}{2}R$, and its molar specific heat at constant pressure ($C_p$) is $\frac{5}{2}R$. The ratio $\gamma = C_p/C_v$ therefore equals $5/3 \approx 1.67$. This value determines how steeply the pressure changes with respect to volume changes during adiabatic transformations compared to isothermal ones.
3. Poisson’s Ratio and Equations
The constant $\gamma$ is known as Poisson’s ratio or the adiabatic index. Aside from the $P-V$ relationship, there are two other useful forms of the adiabatic equation for ideal gases: $TV^{\gamma-1} = \text{constant}$ and $P^{1-\gamma}T^\gamma = \text{constant}$. These allow for the calculation of temperature changes if pressure or volume changes are known. In the problem provided, the steepness of the $P-V$ curve for $\gamma = 5/3$ ensures that a significant reduction in volume (to $1/8$) leads to a much larger increase in pressure (32 times), which is higher than the 8-fold increase that would be seen in an isothermal process.
4. Work Done in Adiabatic Compression
During adiabatic compression, work is done on the gas by an external force. Since no heat leaves the system, all of this work increases the kinetic energy of the atoms, raising the gas temperature. The work done is calculated as $W = \frac{P_1V_1 – P_2V_2}{\gamma – 1}$. In this specific problem, the final pressure is 32 times the initial, and the final volume is 1/8 of the initial. This results in a final state where $P_2V_2 = (32P_1)(V_1/8) = 4P_1V_1$, implying that the final temperature is four times the initial temperature ($T_2 = 4T_1$). This significant temperature jump is characteristic of high-ratio adiabatic compressions.
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FAQs
1
Why is the process considered adiabatic?
Because the container is thermally insulated (no heat transfer) and the compression is “sudden” (no time for heat transfer).
2
What would be the answer if the process was isothermal?
In an isothermal process ($PV=constant$), if volume becomes $1/8$, pressure would become $8$ times.
3
How do we know $\gamma = 5/3$?
The question explicitly states it, and it’s the standard value for a monoatomic gas.
4
Does the mass of the gas affect the ratio?
No, the ratio of pressures depends only on the volume ratio and the adiabatic index $\gamma$.
5
What is the value of $\gamma$ for a diatomic gas?
For a diatomic gas like $O_2$ or $N_2$, $\gamma$ is typically $7/5$ or $1.4$.
6
How to calculate $(8)^{5/3}$ without a calculator?
Find the cube root of 8 first ($\sqrt[3]{8}=2$), then raise it to the power of 5 ($2^5 = 32$).
7
Can this happen in real-life engines?
Yes, the compression stroke in internal combustion engines is often modeled as adiabatic because it happens very quickly.
8
Is an adiabatic process reversible?
An adiabatic process is reversible if it is carried out quasi-statically (very slowly). If done “suddenly”, it is usually irreversible.
9
What happens to entropy in this process?
In a reversible adiabatic process, entropy is constant. In a “sudden” (irreversible) adiabatic process, entropy increases.
10
What is the temperature ratio for this problem?
Using $TV^{\gamma-1} = \text{const}$, $T_2/T_1 = (V_1/V_2)^{2/3} = 8^{2/3} = 4$. The temperature quadruples.
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