Question

An ideal gas at $27^\circ\text{C}$ is compressed adiabatically to $\left(\frac{8}{27}\right)$ of its original volume. If the ratio of specific heats, $\gamma = \frac{5}{3}$ then the rise in temperature of the gas is

• A. $500\text{ K}$
• B. $125\text{ K}$
• C. $250\text{ K}$
• D. $375\text{ K}$

Hint:

Always read carefully to see whether a question asks for the final temperature ($675\text{ K}$) or the rise in temperature ($675 - 300 = 375\text{ K}$). Under time pressure, it is easy to accidentally pick option (A) because $500\text{ K}$ looks like a round number, or misread the final temperature value!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
An ideal gas sample has an initial temperature of $T_1 = 27^\circ\text{C}$. It undergoes a rapid adiabatic compression such that its final volume shrinks to a fraction of its original value ($V_2 = \frac{8}{27}V_1$). Given the adiabatic index $\gamma = \frac{5}{3}$, we need to calculate the total temperature increase ($\Delta T = T_2 - T_1$).

Step 2: Key Formula or Approach:
For any ideal gas system undergoing an adiabatic process, the relationship connecting temperature and volume variables is: $$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \implies \frac{T_2}{T_1} = \left( \frac{V_1}{V_2} \right)^{\gamma-1}$$ We must convert the initial temperature from Celsius to Kelvin ($T(\text{K}) = T(^\circ\text{C}) + 273.15$) before applying this proportional law.

Step 3: Detailed Explanation:
First, convert the initial temperature $T_1$ into Kelvin: $$T_1 = 27^\circ\text{C} + 273 = 300\text{ K}$$ The problem states that the volume ratio is $\frac{V_2}{V_1} = \frac{8}{27}$, which means its inverse is $\frac{V_1}{V_2} = \frac{27}{8}$. Now, compute the exponential power term $\gamma - 1$: $$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$$ Substitute these values into our adiabatic ratio equation to solve for $T_2$: $$\frac{T_2}{300} = \left( \frac{27}{8} \right)^{\frac{2}{3}}$$ Simplify the fractional term inside the parentheses by breaking it down into cubics: $\frac{27}{8} = \left(\frac{3}{2}\right)^3$. $$\frac{T_2}{300} = \left[ \left(\frac{3}{2}\right)^3 \right]^{\frac{2}{3}} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$$ Isolate $T_2$: $$T_2 = 300 \times \frac{9}{4} = 75 \times 9 = 675\text{ K}$$ The question asks for the

rise in temperature ($\Delta T$), not the final absolute temperature: $$\Delta T = T_2 - T_1 = 675\text{ K} - 300\text{ K} = 375\text{ K}$$ This matches option (D).

Step 4: Final Answer:
The net rise in the temperature of the gas is $375\text{ K}$, which corresponds to option (D).