The piston cylinder arrangement shown contains a diatomic gas at temperature 300 K. The cross-sectio

Q: Why is the first process isobaric?

The piston is free to move and the external atmospheric pressure plus the weight of the piston remains constant. Thus, the gas expands at a constant pressure.

Q: Why do we use $\gamma = 1.4$?

The problem specifies a diatomic gas. For diatomic gases at normal temperatures, the degrees of freedom are 5 (3 translational + 2 rotational), which gives $\gamma = 1 + 2/f = 1 + 2/5 = 1.4$.

Q: Can we use the ideal gas law directly for the second part?

You could use $PV = nRT$, but you don't know the final pressure. The adiabatic relation $TV^{\gamma-1} = \text{constant}$ directly connects the known volumes (heights) to the temperatures without needing the pressure.

Q: What does 'brought back to its original height' mean for the volume?

It means the final volume of the gas is exactly the same as the initial volume before the heating process began. So, $V_f = V_1 = A \times 1 \text{ m}$.

Q: Why does the temperature increase during the adiabatic compression?

In an adiabatic compression, work is done ON the gas. Since no heat escapes ($Q=0$), all this work goes into increasing the internal energy of the gas, which manifests as a rise in temperature.

Question

The piston cylinder arrangement shown contains a diatomic gas at temperature $300 	ext{ K}$. The cross-sectional area of the cylinder is $1 	ext{ m}^2$. Initially the height of the piston above the base of the cylinder is $1 	ext{ m}$. The temperature is now raised to $400 	ext{ K}$ at constant pressure. Find the new height of the piston above the base of the cylinder.
If the piston is now brought back to its original height without any heat loss, find the new equilibrium temperature of the gas. You can leave the answer in fraction.

Sigma Percentile · Accepted Answer

The problem presents a classic two-stage thermodynamic journey of a diatomic gas trapped in a cylinder. It tests our understanding of isobaric (constant pressure) and adiabatic (no heat exchange) processes. Let's break down the physics behind each step.

### Analyzing the Setup
We start with a cylinder containing a diatomic gas at an initial temperature $T_1 = 300 	ext{ K}$. The piston is at a height $h_1 = 1 	ext{ m}$, and the cross-sectional area is $A = 1 	ext{ m}^2$.

The first event is heating the gas to $T_2 = 400 	ext{ K}$ at constant pressure. Why is the pressure constant? Because the piston is free to move, and the downward forces (atmospheric pressure and the piston's weight) don't change.

### The Isobaric Expansion
According to Charles's Law, at constant pressure, the volume of an ideal gas is directly proportional to its absolute temperature ($V \propto T$).

Since the volume of the cylinder is $V = A \cdot h$, and the area $A$ is constant, the height of the piston is directly proportional to the temperature:

$$ \frac{h_2}{h_1} = \frac{T_2}{T_1} $$

Substituting our known values:

$$ h_2 = 1 \cdot \frac{400}{300} = \frac{4}{3} 	ext{ m} $$

So, the piston rises to a new height of $\frac{4}{3} 	ext{ m}$.

### The Adiabatic Compression
Next, the piston is pushed back down to its original height of $1 	ext{ m}$. The crucial phrase here is **"without any heat loss"**, which is the hallmark of an adiabatic process ($Q = 0$).

For an adiabatic process, the relationship between temperature and volume is governed by:

$$ T \cdot V^{\gamma - 1} = 	ext{constant} $$

For a diatomic gas, the heat capacity ratio $\gamma$ is $1.4$. 
Let's set up the equation connecting the state before compression (let's call it state $i$) and after compression (state $f$):

$$ T_f \cdot V_f^{\gamma - 1} = T_i \cdot V_i^{\gamma - 1} $$

Again, replacing volume with $A \cdot h$:

$$ T_f \cdot h_f^{\gamma - 1} = T_i \cdot h_i^{\gamma - 1} $$

### Final Calculation
We know the initial state for this compression is the end state of the previous expansion: $T_i = 400 	ext{ K}$ and $h_i = \frac{4}{3} 	ext{ m}$. The final height is $h_f = 1 	ext{ m}$.

$$ T_f \cdot (1)^{1.4 - 1} = 400 \cdot \left(\frac{4}{3}ight)^{1.4 - 1} $$

$$ T_f = 400 \cdot \left(\frac{4}{3}ight)^{0.4} $$

Calculating this value gives:

$$ T_f \approx 448.8 	ext{ K} $$

**Conclusion:** The gas undergoes an expansion followed by a compression. Because the compression was adiabatic, the work done on the gas increased its internal energy, leaving it at a higher temperature than it was before the compression started.

Visualized Solution

$T_{1} = 300 K, h_{1} = 1 m$

$P = constant ⟹ V \propto T$

$V = A \cdot h ⟹ h \propto T$

$h_{2} = h_{1} \cdot \frac{T _{2}}{T _{1}}$

$h_{2} = \frac{4}{3} m$

$Q = 0$

$T \cdot V^{γ - 1} = constant$

$T_{f} \cdot h_{f}^{γ - 1} = T_{i} \cdot h_{i}^{γ - 1}$

$T_{f} = 400 \cdot (\frac{4/3}{1})^{1.4 - 1}$

$T_{f} \approx 448.8 K$

$h_{2} = \frac{4}{3} m, T_{f} = 448.8 K$

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup We start with a cylinder containing a diatomic gas at an initial temperature $T_{1} = 300 K$

The Isobaric Expansion

The Adiabatic Compression Next, the piston is pushed back down to its original height of $1 m$

Final Calculation We know the initial state for this compression is the end state of the previous expansion: $T_{i} = 400 K$ and $h_{i} = \frac{4}{3} m$

Similar Questions

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

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​=300 K,h1​=1 m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P=constant⟹V∝T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=A⋅h⟹h∝T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h2​=h1​⋅T1​T2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h2​=34​ m

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T⋅Vγ−1=constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tf​⋅hfγ−1​=Ti​⋅hiγ−1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tf​=400⋅(14/3​)1.4−1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tf​≈448.8 K

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h2​=34​ m,Tf​=448.8 K