An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in

Q: Why is the initial pressure equal to the atmospheric pressure?

Initially, the spring is in its relaxed state, meaning it exerts no force on the piston. For the piston to be in equilibrium, the force from the gas inside must perfectly balance the force from the atmosphere outside. Hence, $p_1 A = p_{atm} A$, which gives $p_1 = p_{atm}$.

Q: How do we know the process is a straight line on the p-V diagram?

The pressure of the gas is given by $p = p_{atm} + \frac{kx}{A}$. Since the displacement $x$ is related to the volume change by $x = \frac{V - V_1}{A}$, we can write $p = p_1 + \frac{k}{A^2}(V - V_1)$. This is a linear equation of the form $p = mV + c$, which represents a straight line on the $p-V$ diagram.

Q: Why do we use $C_V = \frac{3}{2}R$ for the internal energy calculation?

The problem states that the gas is an ideal monoatomic gas. For monoatomic gases, the degrees of freedom $f = 3$. The molar heat capacity at constant volume is $C_V = \frac{f}{2}R = \frac{3}{2}R$. The internal energy is always given by $U = nC_V T = \frac{3}{2}pV$.

Q: Can we use $W = p\Delta V$ to find the work done?

No, you cannot use $W = p\Delta V$ directly because the pressure is not constant during the expansion. As the gas expands, it compresses the spring, which increases the opposing force, causing the gas pressure to rise. You must integrate $p dV$ or use the area of the trapezium on the $p-V$ graph.

Q: Is the heat supplied calculated at constant pressure or constant volume?

Neither! The process is polytropic (specifically, a straight line on the $p-V$ graph), so it's not isobaric or isochoric. To find the heat supplied, you must use the First Law of Thermodynamics: $\Delta Q = W + \Delta U$, calculating the work done and the change in internal energy separately.

Question

An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature $T_1$, pressure $p_1$ and volume $V_1$ and the spring is in its relaxed state. The gas is then heated very slowly to temperature $T_2$, pressure $p_2$ and volume $V_2$. During this process the piston moves out by a distance $x$. Ignoring the friction between the piston and the cylinder, the correct statements is/are

Options:

(A) If $V_2 = 2V_1$ and $T_2 = 3T_1$, then the energy stored in the spring is $\frac{1}{4} p_1V_1$

(B) If $V_2 = 2V_1$ and $T_2 = 3T_1$, then the change in internal energy is $3p_1V_1$

(C) If $V_2 = 3V_1$ and $T_2 = 4T_1$, then the work done by the gas is $\frac{7}{3} p_1V_1$

(D) If $V_2 = 3V_1$ and $T_2 = 4T_1$, then the heat supplied to the gas is $\frac{17}{6} p_1V_1$

Sigma Percentile · Accepted Answer

Imagine you are standing next to a transparent cylinder. Inside, a monoatomic gas is quietly resting, confined by a piston. Attached to the outside of this piston is a spring, currently in its relaxed state. This is our starting point, a perfect equilibrium.

### Analyzing the Setup

Because the spring is relaxed, it exerts absolutely no force on the piston. For the piston to remain stationary, the pressure of the gas inside must perfectly balance the atmospheric pressure outside. Therefore, our initial condition is beautifully simple: **$p_1 = p_{atm}$**.

Now, let's turn up the heat. As the gas warms up, it expands, pushing the piston outward by a distance $x$. But now, the gas has to fight two battles: it must push against the constant atmospheric pressure, AND it must compress the spring. The force balance on the piston becomes:

$$p_2 A = p_{atm} A + kx$$

Since we know $p_{atm} = p_1$, we can rewrite this as our master equation for the pressure:

$$p_2 = p_1 + \frac{kx}{A}$$

Simultaneously, the volume of the gas has increased by the volume of the cylinder swept by the piston, which is $Ax$. So, the new volume is $V_2 = V_1 + Ax$. This gives us a direct way to find the displacement: **$x = \frac{V_2 - V_1}{A}$**.

### Evaluating Options (a) and (b)

Let's test the first scenario where the volume doubles ($V_2 = 2V_1$) and the temperature triples ($T_2 = 3T_1$). We can find the new pressure using the ideal gas law:

$$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2} \Rightarrow p_2 = p_1 \left(\frac{T_2}{T_1}\right) \left(\frac{V_1}{V_2}\right)$$

Substituting our values, $p_2 = p_1 (3) \left(\frac{1}{2}\right) = \frac{3}{2} p_1$.

Now, let's find the energy stored in the spring. The displacement is $x = \frac{2V_1 - V_1}{A} = \frac{V_1}{A}$. From our master pressure equation, the spring force term is $\frac{kx}{A} = p_2 - p_1 = \frac{3}{2}p_1 - p_1 = \frac{1}{2}p_1$. This means $kx = \frac{1}{2}p_1 A$.

The potential energy of the spring is $U_s = \frac{1}{2}kx^2$. Substituting what we found:

$$U_s = \frac{1}{2} \left(\frac{p_1 A}{2}\right) \left(\frac{V_1}{A}\right) = \mathbf{\frac{1}{4}p_1 V_1}$$

This perfectly matches option (a)!

What about the internal energy? For an ideal monoatomic gas, the change in internal energy is $\Delta U = \frac{3}{2} (p_2 V_2 - p_1 V_1)$. Let's plug in the final state:

$$\Delta U = \frac{3}{2} \left[ \left(\frac{3}{2}p_1\right)(2V_1) - p_1 V_1 \right] = \frac{3}{2} [3p_1 V_1 - p_1 V_1] = \mathbf{3p_1 V_1}$$

Option (b) is also correct!

### Evaluating Options (c) and (d)

Now let's look at the second scenario: the volume triples ($V_2 = 3V_1$) and the temperature quadruples ($T_2 = 4T_1$). Using the ideal gas law again, the new pressure is $p_2 = p_1 (4) \left(\frac{1}{3}\right) = \frac{4}{3} p_1$.

The displacement is now $x = \frac{3V_1 - V_1}{A} = \frac{2V_1}{A}$. The spring force term becomes $\frac{kx}{A} = \frac{4}{3}p_1 - p_1 = \frac{1}{3}p_1$, which means $kx = \frac{1}{3}p_1 A$.

To find the work done by the gas, we can calculate the area under the $p-V$ graph. Because the pressure increases linearly with volume (due to the spring), the process forms a straight line, creating a trapezium on the graph. The area is:

$$W = \frac{1}{2}(p_1 + p_2)(V_2 - V_1)$$

$$W = \frac{1}{2} \left(p_1 + \frac{4}{3}p_1\right) (3V_1 - V_1) = \frac{1}{2} \left(\frac{7}{3}p_1\right) (2V_1) = \mathbf{\frac{7}{3}p_1 V_1}$$

Option (c) is correct!

Finally, let's check the heat supplied using the First Law of Thermodynamics: $\Delta Q = W + \Delta U$. We first need the new change in internal energy:

$$\Delta U = \frac{3}{2} \left[ \left(\frac{4}{3}p_1\right)(3V_1) - p_1 V_1 \right] = \frac{3}{2} [4p_1 V_1 - p_1 V_1] = \frac{9}{2} p_1 V_1$$

Adding the work done and the internal energy change:

$$\Delta Q = \frac{7}{3}p_1 V_1 + \frac{9}{2}p_1 V_1 = \left(\frac{14 + 27}{6}\right) p_1 V_1 = \mathbf{\frac{41}{6}p_1 V_1}$$

Option (d) claims the heat supplied is $\frac{17}{6} p_1 V_1$, which is incorrect.

### The Final Verdict

By carefully tracking the forces and applying the fundamental laws of thermodynamics, we've successfully navigated this problem. The correct statements are indeed (a), (b), and (c).

Select Answer:

Visualized Solution

$I ni t ia l S t a t e$

$G e n er a l F ina l S t a t e$

$A na l y z in g O pt i o n s (a) an d (b)$

$S p r in g E n er g y f or O pt i o n (a)$

$I n t er na l E n er g y f or O pt i o n (b)$

$A na l y z in g O pt i o n s (c) an d (d)$

$W or k D o n e f or O pt i o n (c)$

$H e a t S u ppl i e d f or O pt i o n (d)$

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

Evaluating Options (a) and (b)

Evaluating Options (c) and (d)

The Final Verdict

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