Thermodynamic process is shown below on a p-V diagram for one mole of an ideal gas. If V2 = 2V1, the

Q: Why can't we just use the adiabatic relation $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$?

Because this is not necessarily an adiabatic process! The adiabatic relation only holds when $pV^\gamma = \text{constant}$. Here, the exponent is $1/2$, which is a general polytropic process $pV^x = \text{constant}$. You must derive the $T-V$ relation specifically for $x = 1/2$.

Q: What does $pV^{1/2} = \text{constant}$ mean physically?

It represents a polytropic process. Unlike isothermal (constant $T$) or adiabatic (no heat exchange) processes, a polytropic process is a more general case where both heat exchange and work done occur in a specific proportional manner.

Q: How did the power of $V$ change from $1/2$ to $-1/2$?

When we substituted $p = \frac{nRT}{V}$ into $pV^{1/2}$, we got $\frac{nRT}{V} \cdot V^{1/2}$. Mathematically, $\frac{V^{1/2}}{V^1} = V^{1/2 - 1} = V^{-1/2}$.

Q: Does the number of moles ($n$) matter in this calculation?

No, it doesn't. Because $n$ and $R$ are constants, when we write $\frac{nRT}{V} V^{1/2} = C$, we can just say $T V^{-1/2} = \frac{C}{nR}$. Since $\frac{C}{nR}$ is just another constant, $n$ drops out of the ratio calculation entirely.

Q: What if the question asked for the ratio of pressures $\frac{p_1}{p_2}$ instead?

That would be even faster! You would just use the given equation directly: $p_1 V_1^{1/2} = p_2 V_2^{1/2}$, which rearranges to $\frac{p_1}{p_2} = \left(\frac{V_2}{V_1}\right)^{1/2}$. Substituting $V_2 = 2V_1$ would give $\sqrt{2}$.

Question

Thermodynamic process is shown below on a p-V diagram for one mole of an ideal gas.
If $V_2 = 2V_1$, then the ratio of temperature $\frac{T_2}{T_1}$ is

Options:

(A) $1/\sqrt{2}$

(B) $\sqrt{2}$

(C) $1/2$

(D) $2$

Sigma Percentile · Accepted Answer

This problem is a beautiful demonstration of how we can manipulate thermodynamic state variables to find exactly what we need. Let's embark on this journey step-by-step.

### Analyzing the Setup

We are given a $p-V$ diagram showing a gas expanding from state 1 to state 2. The curve is governed by a specific mathematical rule:

$$pV^{1/2} = 	ext{constant}$$

We are also handed a crucial piece of information regarding the volumes: the final volume is exactly twice the initial volume, meaning $V_2 = 2V_1$. Our ultimate goal is to find the ratio of their absolute temperatures, $\frac{T_2}{T_1}$.

### The Master Equation

The process equation relates pressure ($p$) and volume ($V$), but our target involves temperature ($T$). How do we bridge this gap? We call upon the fundamental law of ideal gases:

$$pV = nRT$$

From this, we can isolate pressure:

$$p = \frac{nRT}{V}$$

This simple rearrangement is the key to unlocking the problem.

### Transforming the Process Equation

Now, let's substitute our expression for pressure back into the given process equation. Replacing $p$, we get:

$$\left(\frac{nRT}{V}ight) V^{1/2} = 	ext{constant}$$

Let's simplify the powers of $V$. Dividing $V^{1/2}$ by $V^1$ gives us $V^{-1/2}$. Since the number of moles ($n$) and the universal gas constant ($R$) are constants, we can absorb them into the constant on the right side, creating a new constant, let's call it $C'$.

$$T \cdot V^{-1/2} = C'$$

This is a powerful realization! It tells us that for any two states in this process, the product of temperature and $V^{-1/2}$ remains the same. Therefore:

$$T_1 V_1^{-1/2} = T_2 V_2^{-1/2}$$

### Final Calculation

We are now perfectly positioned to find the required ratio. Let's rearrange our equation to isolate $\frac{T_2}{T_1}$:

$$\frac{T_2}{T_1} = \frac{V_1^{-1/2}}{V_2^{-1/2}}$$

When negative exponents cross the fraction line, they become positive. This elegantly simplifies to:

$$\frac{T_2}{T_1} = \left(\frac{V_2}{V_1}ight)^{1/2}$$

Now, we bring in our initial condition, $V_2 = 2V_1$. Substituting this into our ratio:

$$\frac{T_2}{T_1} = \left(\frac{2V_1}{V_1}ight)^{1/2}$$

The $V_1$ terms cancel out flawlessly, leaving us with:

**$$\frac{T_2}{T_1} = \sqrt{2}$$**

This matches option (b). The beauty of thermodynamics lies in these seamless transformations between state variables!

Animated Solution for Physics - Thermodynamics: Thermodynamic process is shown below on a p-V diagram for one mole of an ideal gas. If $V_{2} = 2 V_{1}$ , then the ratio of temperature $\frac{T _{2}}{T _{1}}$ is

Select Answer:

Visualized Solution

Analyzing the Setup

The Ideal Gas Law

Substituting Pressure

Temperature-Volume Relation

Rearranging for the Ratio

Final Calculation

The Way Forward

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation

Transforming the Process Equation

Final Calculation

Similar Questions

During an experiment, an ideal gas is found to obey an additional law $p^{2} V = constant$ . The gas is initially at a temperature $T$ and volume $V$ . When it expands to a volume $2 V$ , the temperature becomes ……… .

One mole of a monoatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V-T) diagram. The correct statement(s) is/are: [R is the gas constant]

$n$ moles of an ideal gas undergoes a process $A$ and $B$ as shown in the figure. The maximum temperature of the gas during the process will be

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation $V T = k$ , where $k$ is a constant. In this process, the temperature of the gas is increased by $Δ T$ . The amount of heat absorbed by gas is (where, $R$ is gas constant)

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio $C_{p} / C_{V}$ for the gas is

Select Answer:

Visualized Solution

Analyzing the Setup

The Ideal Gas Law

Substituting Pressure

Temperature-Volume Relation

Rearranging for the Ratio

Final Calculation

The Way Forward

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation

Transforming the Process Equation

Final Calculation

Similar Questions

One mole of a monoatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V-T) diagram. The correct statement(s) is/are: [R is the gas constant]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n moles of an ideal gas undergoes a process A and B as shown in the figure. The maximum temperature of the gas during the process will be

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio Cp​/CV​ for the gas is

$n$ moles of an ideal gas undergoes a process $A$ and $B$ as shown in the figure. The maximum temperature of the gas during the process will be

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio $C_{p} / C_{V}$ for the gas is