Twenty cubic centimeters of monatomic gas at 12 ^∘ C and 100 kPa is suddenly (and adiabatically)

step 1 We have to find the value of new pressure and temperature for adiatic process. Some given values

Twenty cubic centimeters of monatomic gas at 12 ^∘ C and 100...

Twenty cubic centimeters of monatomic gas at $12{ }^{\circ} \mathrm{C}$ and 100 $\mathrm{kPa}$ is suddenly (and adiabatically) compressed to $0.50 \mathrm{~cm}^{3}$. Assume that we are dealing with an ideal gas. What are its new pressure and temperature? For an adiabatic change involving an ideal gas, $P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}$ where $\mathrm{Y}=1.67$ for a monatomic gas. Hence, $$P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=\left(1.00 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(\frac{20}{0.50}\right)^{1.67}=4.74 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}=47 \mathrm{MPa}$$ To find the final temperature, we could use $P_{1} V_{1} / T_{1}=P_{2} V_{2} / T_{2}$. Instead, let us use $$T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma}$$ or $$T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}=(285 \mathrm{~K})\left(\frac{20}{0.50}\right)^{0.67}=(285 \mathrm{~K})(11.8)=3.4 \times 10^{3} \mathrm{~K}$$ As a check, $$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$$ $$ \begin{aligned} \frac{\left(1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(20 \mathrm{~cm}^{3}\right)}{285 \mathrm{~K}} &=\frac{\left(4.74 \times 107 \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.50 \mathrm{~cm}^{3}\right)}{3370 \mathrm{~K}} \\ 7000 &=7000 \end{aligned} $$

Twenty cubic centimeters of monatomic gas at $12{ }^{\circ} \mathrm{C}$ and 100 $\mathrm{kPa}$ is suddenly (and adiabatically) compressed to $0.50 \mathrm{~cm}^{3}$. Assume that we are dealing with an ideal gas. What are its new pressure and temperature?
For an adiabatic change involving an ideal gas, $P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}$ where $\mathrm{Y}=1.67$ for a monatomic gas. Hence,
$$P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=\left(1.00 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(\frac{20}{0.50}\right)^{1.67}=4.74 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}=47 \mathrm{MPa}$$
To find the final temperature, we could use $P_{1} V_{1} / T_{1}=P_{2} V_{2} / T_{2}$. Instead, let us use
$$T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma}$$
or
$$T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}=(285 \mathrm{~K})\left(\frac{20}{0.50}\right)^{0.67}=(285 \mathrm{~K})(11.8)=3.4 \times 10^{3} \mathrm{~K}$$
As a check,
$$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$$
$$
\begin{aligned}
\frac{\left(1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(20 \mathrm{~cm}^{3}\right)}{285 \mathrm{~K}} &=\frac{\left(4.74 \times 107 \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.50 \mathrm{~cm}^{3}\right)}{3370 \mathrm{~K}} \\
7000 &=7000
\end{aligned}
$$

Key Concepts

Temperature-Volume Relationship in Adiabatic Processes

In an adiabatic process for an ideal gas, there exists a relationship between temperature and volume, often written as T V^(?-1) = constant. This connection allows one to determine the change in temperature of the gas when its volume changes adiabatically, complementing the pressure-volume relationship provided by P V^? = constant.

Adiabatic Index (?)

The adiabatic index, denoted by ?, is the ratio of specific heats at constant pressure and constant volume (? = Cp/Cv). Its value depends on the nature of the gas (for instance, ? = 1.67 for a monatomic gas) and plays a key role in determining how the pressure, volume, and temperature change during an adiabatic process.

Adiabatic Process

An adiabatic process is one in which no heat is exchanged with the surroundings, meaning that all changes in the internal energy of the system result solely from the work done on or by the system. This concept is crucial in thermodynamics when analyzing rapid processes where there is insufficient time for heat transfer, making energy changes attributable entirely to work.

Adiabatic Equation (P V^? = constant)

For an ideal gas undergoing an adiabatic process, the product of the pressure and the volume raised to the power of the adiabatic index (?) remains constant. This relation, P V^? = constant, provides a direct method to relate the initial and final pressures and volumes of the gas during a rapid, no-heat-transfer transformation.

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, relates the pressure, volume, and temperature of an ideal gas via the amount of substance and the universal gas constant. This law is foundational in connecting the state variables and serves as the basis for deriving other relations used in adiabatic processes.

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