Starting at temperature 300 K, one mole of an ideal diatomic gas (gamma = 1.4) is first compressed a

Q: Why do we use $TV^{\gamma-1}$ instead of $PV^\gamma$?

Because we are given initial temperature and volumes, and asked for final temperature. Using $TV^{\gamma-1}$ directly connects the knowns to the unknowns, saving you the extra step of calculating intermediate pressures.

Q: How do I calculate $16^{0.4}$ without a calculator?

The trick is to express the base as a power of a smaller prime. Recognize that $16 = 2^4$. Then, $(2^4)^{0.4} = 2^{4 \times 0.4} = 2^{1.6}$. You can keep it in this exponent form until the very final calculation.

Q: Why does the temperature increase during adiabatic compression?

During compression, work is done *on* the gas. Since the process is adiabatic, no heat can escape ($Q=0$). According to the First Law of Thermodynamics ($\Delta U = Q - W$), the work done on the gas entirely goes into increasing its internal energy, which manifests as a rise in temperature.

Q: Why is the final volume $V_1/8$ and not $2V_1$?

Read the question carefully! It states the gas expands to $2V_2$, not $2V_1$. Since the intermediate volume $V_2$ is $V_1/16$, the final volume is $2 \times (V_1/16) = V_1/8$.

Q: Does the gas absorb heat during the isobaric process?

Yes, absolutely. To expand while keeping the pressure constant, the gas must do work against its surroundings. To maintain its internal energy and do this work simultaneously, heat must be continuously added to the system.

Question

Starting at temperature $300	ext{ K}$, one mole of an ideal diatomic gas ($\gamma = 1.4$) is first compressed adiabatically from volume $V_1$ to $V_2 = \frac{V_1}{16}$. It is then allowed to expand isobarically to volume $2V_2$. If all the processes are the quasi-static, then the final temperature of the gas (in $^\circ	ext{K}$) is (to the nearest integer) ......... .

Sigma Percentile · Accepted Answer

### Visualizing the Thermodynamic Path

Imagine you are tracking the journey of a diatomic gas trapped inside a cylinder. The problem describes a two-part journey, which is best visualized on a $p-V$ (pressure-volume) diagram.

First, the gas undergoes an **adiabatic compression**. This means the piston is pushed down so rapidly that no heat can escape the system ($Q=0$). As the volume shrinks from $V_1$ to $V_1/16$, the pressure spikes dramatically, and the internal energy (and thus temperature) shoots up.

Next, the gas is allowed to undergo an **isobaric expansion**. The gas pushes the piston back up, but this time, heat is added to the system to keep the pressure perfectly constant. The volume doubles from its intermediate state, reaching $2V_2$. Our goal is to find the final temperature at the end of this entire sequence.

### Phase 1: The Adiabatic Squeeze

To find the temperature after the first process, we need the adiabatic relationship between temperature and volume. For an ideal gas undergoing a reversible adiabatic process, the product of temperature and volume raised to the power of $\gamma - 1$ remains constant:

$$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$$

We are given the initial temperature $T_1 = 300	ext{ K}$, the specific heat ratio $\gamma = 1.4$, and the final volume $V_2 = \frac{V_1}{16}$. Let's substitute these values into our master equation:

$$300 \cdot V_1^{1.4-1} = T_2 \cdot \left(\frac{V_1}{16}ight)^{1.4-1}$$

Simplifying the exponents, we get:

$$300 \cdot V_1^{0.4} = T_2 \cdot \frac{V_1^{0.4}}{16^{0.4}}$$

Notice how beautifully the $V_1^{0.4}$ terms cancel out from both sides. This leaves us with an equation purely in terms of $T_2$:

$$T_2 = 300 \cdot 16^{0.4}$$

To solve this without a calculator, recognize that $16 = 2^4$. Therefore, $16^{0.4} = (2^4)^{0.4} = 2^{1.6}$. So, the intermediate temperature is:

$$T_2 = 300 \cdot 2^{1.6}	ext{ K}$$

### Phase 2: The Isobaric Stretch

Now the gas enters the second phase: an isobaric expansion. Isobaric means the pressure remains constant. According to **Charles's Law**, at a constant pressure, the volume of an ideal gas is directly proportional to its absolute temperature:

$$\frac{V_2}{T_2} = \frac{V_3}{T_3}$$

The problem states that the gas expands to a volume $V_3 = 2V_2$. Let's plug this into Charles's Law:

$$\frac{V_2}{300 \cdot 2^{1.6}} = \frac{2V_2}{T_3}$$

Once again, the volume terms ($V_2$) cancel out neatly. Solving for the final temperature $T_3$, we find that it is simply twice the intermediate temperature:

$$T_3 = 2 \cdot 300 \cdot 2^{1.6}$$

Using the laws of exponents, we add the powers of base 2 ($2^1 \cdot 2^{1.6} = 2^{2.6}$):

$$T_3 = 300 \cdot 2^{2.6}	ext{ K}$$

### The Final Destination

All that remains is to evaluate this final expression. We can break down $2^{2.6}$ as $2^2 \cdot 2^{0.6} = 4 \cdot 2^{3/5}$.

Knowing that $2^3 = 8$, we need the fifth root of 8, which is approximately $1.5157$. Multiplying this by 4 gives us $6.0628$.

Finally, multiplying by the initial 300 K:

$$T_3 \approx 300 \cdot 6.0628 \approx 1818.85	ext{ K}$$

Rounding to the nearest integer as requested by the problem, we arrive at our final answer:

**Final Temperature = 1819 K**

Enter Numerical Value:

Visualized Solution

$p - V Diagram$

$Adiabatic Relation$

$Substitution$

$Calculating T_{2}$

$Isobaric Relation$

$Substitution$

$Calculating T_{3}$

$Final Answer$

The Sigma Insight: Thermodynamic Processes

Visualizing the Thermodynamic Path

Phase 1

Phase 2

The Final Destination

Similar Questions

Two moles of an ideal monoatomic gas occupies a volume $V$ at $2 7^{\circ} C$ . The gas expands adiabatically to a volume $2 V$ . Calculate (i) the final temperature of the gas and (ii) change in its internal energy.

A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_{i}$ (in kelvin) and the final temperature is $a T_{i}$ , the value of $a$ is

At $2 7^{\circ} C$ two moles of an ideal monoatomic gas occupy a volume $V$ . The gas expands adiabatically to a volume $2 V$ . Calculate (a) the final temperature of the gas, (b) change in its internal energy, (c) the work done by the gas during this process.

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p−V Diagram

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Adiabatic Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating T2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isobaric Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating T3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

The Sigma Insight: Thermodynamic Processes

Visualizing the Thermodynamic Path

Phase 1

Phase 2

The Final Destination

Similar Questions

$p - V Diagram$

$Adiabatic Relation$

$Substitution$

$Calculating T_{2}$

$Isobaric Relation$

$Substitution$

$Calculating T_{3}$

$Final Answer$