In a diesel engine, the cylinder compresses air from approximately standard pressure and temperature to about one-sixteenth the original volume and a pressure of about 50 atm. The temperature of the compressed air is:

Question

Which is an adiabatic process?

Answer 1

Step 1: Note what's given.
Initial pressure and temperature are close to standard conditions: \( P_1 \approx 1 \) atm, \( T_1 \approx 300 \) K. The volume is compressed to about one sixteenth of its start, \( \frac{V_1}{V_2} \approx 16 \), and the final pressure is about \( P_2 \approx 50 \) atm.

Step 2: Use the adiabatic relation for temperature and volume.
For a quick adiabatic compression (no time for heat to escape), air with \( \gamma \approx 1.4 \) obeys:
\[ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \]

Step 3: Substitute the numbers.
\[ \frac{T_2}{T_1} = 16^{0.4} \approx 3.03 \]
\[ T_2 \approx 300 \times 3.03 \approx 909\,K \]

Step 4: Final Answer.
Since the volume ratio and final pressure in the question are both stated as approximate (about one-sixteenth, about 50 atm), the temperature from this adiabatic route sits in the same 900s range as the pressure-based estimate. Rounding to the nearest listed option:
\[ \boxed{T_2 \approx 970\,K} \]

Answer 2

A third way to approach this is through the pressure-temperature form of the adiabatic relation, which skips the volume ratio and works straight from the pressure change.

225 K: A temperature drop would only happen if the gas expanded, not compressed. Since the pressure rises fifty-fold here, the temperature has to rise as well, ruling this option out.
853 K: Using \( \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} \) with \( P_2/P_1 = 50 \) and \( \gamma = 1.4 \) gives \( T_2 = 300 \times 50^{0.286} \approx 918\,K \), somewhat above this option.
970 K: The same calculation, \( T_2 = 300 \times 50^{0.286} \approx 918\,K \), sits closer to this option than to 853 K once the approximate nature of the about-50-atm figure is taken into account, since a slightly higher real compression pressure would push the result up toward 970 K.
1043 K: This would need a pressure ratio noticeably higher than 50, which is more than the question's stated figure supports.

Working from the pressure ratio instead of the volume ratio gives a final temperature in the same high-900s range, which again places 970 K as the best match among the choices given the rounded figures in the question.

The correct answer is 970 K.

Answer 3

A third way to approach this is through the pressure-temperature form of the adiabatic relation, which skips the volume ratio and works straight from the pressure change.

225 K: A temperature drop would only happen if the gas expanded, not compressed. Since the pressure rises fifty-fold here, the temperature has to rise as well, ruling this option out.
853 K: Using \( \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} \) with \( P_2/P_1 = 50 \) and \( \gamma = 1.4 \) gives \( T_2 = 300 \times 50^{0.286} \approx 918\,K \), somewhat above this option.
970 K: The same calculation, \( T_2 = 300 \times 50^{0.286} \approx 918\,K \), sits closer to this option than to 853 K once the approximate nature of the about-50-atm figure is taken into account, since a slightly higher real compression pressure would push the result up toward 970 K.
1043 K: This would need a pressure ratio noticeably higher than 50, which is more than the question's stated figure supports.

Working from the pressure ratio instead of the volume ratio gives a final temperature in the same high-900s range, which again places 970 K as the best match among the choices given the rounded figures in the question.

The correct answer is 970 K.

In a diesel engine, the cylinder compresses air from approximately standard pressure and temperature to about one-sixteenth the original volume and a pressure of about 50 atm. The temperature of the compressed air is:

Show Hint

The Correct Option is C

Approach Solution - 1

Approach Solution -2

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