An ideal monoatomic gas is confined in a cylinder by a spring-loaded piston of cross-section 8.0 x 1

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

Because the piston is in equilibrium and the spring is in its relaxed state, meaning it exerts no force. The only forces on the piston are the gas pressure from the inside and the atmospheric pressure from the outside.

Q: Why do we add the spring force to the atmospheric pressure to find the final pressure?

In the final state, the gas is pushing outwards, while both the atmosphere and the compressed spring are pushing inwards. For the piston to be in equilibrium, the outward force ($p_f A$) must equal the sum of the inward forces ($p_0 A + kx$).

Q: Can I use $W = p \Delta V$ for the entire work done?

No, because the pressure of the gas is not constant during the expansion. The force exerted by the spring increases as it gets compressed. You must calculate the work done against the atmosphere ($p_0 \Delta V$) and the work done against the spring ($\frac{1}{2} k x^2$) separately.

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

The problem states that the gas is monoatomic. For an ideal monoatomic gas, the only degrees of freedom are the three translational modes, which gives a molar heat capacity at constant volume of $\frac{3}{2} R$.

Q: What happens to the heat capacity of the heater coil?

The problem explicitly states to neglect the heat capacity of the heater coil and heat loss through the wires. This means we assume 100% of the electrical energy supplied by the heater goes directly into the gas as heat $Q$.

Question

An ideal monoatomic gas is confined in a cylinder by a spring-loaded piston of cross-section $8.0 	imes 10^{-3} 	ext{ m}^2$. Initially the gas is at $300 	ext{ K}$ and occupies a volume of $2.4 	imes 10^{-3} 	ext{ m}^3$ and the spring is in its relaxed (unstretched, uncompressed) state. The gas is heated by a small electric heater until the piston moves out slowly by $0.1 	ext{ m}$.
Calculate the final temperature of the gas and the heat supplied (in joules) by the heater. The force constant of the spring is $8000 	ext{ N/m}$, and the atmospheric pressure $1.0 	imes 10^5 	ext{ N m}^{-2}$. The cylinder and the piston are thermally insulated. The piston is massless and there is no friction between the piston and the cylinder. Neglect heat loss through the lead wires of the heater. The heat capacity of the heater coil is negligible. Assume the spring to be massless.

Sigma Percentile · Accepted Answer

The beauty of thermodynamics lies in its ability to track every single Joule of energy as it transforms and moves through a system. In this classic problem, we are presented with a fascinating interplay of thermal energy, mechanical work, and elastic potential energy. Let's break down the journey of this ideal monoatomic gas step by step.

### Analyzing the Initial State

Before we turn on the heater, we must establish our baseline. The gas is trapped inside a cylinder by a piston. Crucially, the problem states that the spring is initially in its **relaxed state**. This means the spring exerts absolutely zero force on the piston.

Because the piston is in equilibrium, the forces on it must balance. The only forces acting horizontally are the pressure of the gas pushing outwards and the atmospheric pressure pushing inwards. Therefore, the initial pressure of the gas, $p_i$, is exactly equal to the atmospheric pressure, $p_0$.

$$p_i = p_0 = 1.0 	imes 10^5 	ext{ N/m}^2$$

We are also given the initial volume $V_i = 2.4 	imes 10^{-3} 	ext{ m}^3$ and the initial temperature $T_i = 300 	ext{ K}$.

### The Expansion

When the electric heater is turned on, it pumps thermal energy into the gas. The gas molecules gain kinetic energy, the pressure tries to rise, and the gas expands, pushing the piston outwards. The piston moves slowly by a distance $x = 0.1 	ext{ m}$.

We can easily calculate the new volume. The increase in volume, $\Delta V$, is simply the cross-sectional area of the piston multiplied by the distance it moved:

$$\Delta V = A x = (8.0 	imes 10^{-3} 	ext{ m}^2)(0.1 	ext{ m}) = 0.8 	imes 10^{-3} 	ext{ m}^3$$

The final volume $V_f$ is then:

$$V_f = V_i + \Delta V = 2.4 	imes 10^{-3} + 0.8 	imes 10^{-3} = 3.2 	imes 10^{-3} 	ext{ m}^3$$

### Force Equilibrium

Now, let's look at the piston in its final position. It has moved outwards, compressing the spring. The spring now fights back with a restoring force given by Hooke's Law, $F = kx$.

For the piston to be in equilibrium in this new position, the outward force exerted by the gas must balance **both** the inward force from the atmosphere and the inward force from the compressed spring.

$$p_f A = p_0 A + kx$$

Dividing by the area $A$, we get the final pressure of the gas:

$$p_f = p_0 + \frac{kx}{A}$$

Substituting the given values:

$$p_f = 1.0 	imes 10^5 + \frac{8000 	imes 0.1}{8.0 	imes 10^{-3}} = 1.0 	imes 10^5 + 1.0 	imes 10^5 = 2.0 	imes 10^5 	ext{ N/m}^2$$

Notice how the act of compressing the spring directly forces the gas to reach a higher pressure.

### The Master Equation

With the initial and final states well-defined in terms of pressure and volume, we can find the final temperature using the ideal gas law. Since the cylinder is sealed, the number of moles $n$ remains constant. Therefore, we can use the combined gas law:

$$\frac{p_i V_i}{T_i} = \frac{p_f V_f}{T_f}$$

Rearranging for $T_f$:

$$T_f = T_i \left( \frac{p_f}{p_i} ight) \left( \frac{V_f}{V_i} ight)$$

$$T_f = 300 	imes \left( \frac{2.0 	imes 10^5}{1.0 	imes 10^5} ight) 	imes \left( \frac{3.2 	imes 10^{-3}}{2.4 	imes 10^{-3}} ight) = 300 	imes 2 	imes \frac{4}{3} = 800 	ext{ K}$$

This is the first part of our answer. The gas has heated up significantly!

### The Energy Split

To find the total heat supplied by the heater, we need to invoke the **First Law of Thermodynamics**: $Q = \Delta U + W$. Let's calculate the work done, $W$, first.

The gas does work against two distinct entities: the constant atmospheric pressure and the variable spring force.

1. **Work against the atmosphere:** Since atmospheric pressure is constant, this is simply $W_{	ext{atm}} = p_0 \Delta V$.
2. **Work against the spring:** The force of the spring increases linearly from $0$ to $kx$. The work done is the area under the force-displacement graph, which is the elastic potential energy stored in the spring: $W_{	ext{spring}} = \frac{1}{2} k x^2$.

Total work done:

$$W = p_0 \Delta V + \frac{1}{2} k x^2$$

$$W = (1.0 	imes 10^5)(0.8 	imes 10^{-3}) + \frac{1}{2}(8000)(0.1)^2 = 80 + 40 = 120 	ext{ J}$$

### Internal Energy

Next, we calculate the change in internal energy, $\Delta U$. The problem specifies that we are dealing with a **monoatomic** gas. This is a critical piece of information because it tells us the molar heat capacity at constant volume, $C_V = \frac{3}{2} R$.

$$\Delta U = n C_V \Delta T = n \left( \frac{3}{2} R ight) (T_f - T_i)$$

We don't know $n$ directly, but we can easily find the product $nR$ from the initial state using the ideal gas law, $p_i V_i = n R T_i$:

$$n R = \frac{p_i V_i}{T_i} = \frac{(1.0 	imes 10^5)(2.4 	imes 10^{-3})}{300} = 0.8 	ext{ J/K}$$

Now, substitute this back into the $\Delta U$ equation:

$$\Delta U = \frac{3}{2} (n R) (T_f - T_i) = \frac{3}{2} (0.8)(800 - 300) = 1.2 	imes 500 = 600 	ext{ J}$$

### The First Law

Finally, we bring it all together. The total heat $Q$ supplied by the heater is the sum of the energy that went into heating the gas (increasing its internal energy) and the energy that went into doing mechanical work.

$$Q = \Delta U + W = 600 	ext{ J} + 120 	ext{ J} = 720 	ext{ J}$$

And there we have it! A beautiful demonstration of energy conservation, where electrical energy from the heater is perfectly accounted for in the thermal agitation of the gas molecules, the pushing away of the atmosphere, and the tension of a compressed spring.

Visualized Solution

$I ni t ia l S t a t e o f t h e G a s$

$C a l c u l a t in g F ina l V o l u m e$

$C a l c u l a t in g F ina l P r ess u r e$

$C a l c u l a t in g F ina l T e m p er a t u r e$

$C a l c u l a t in g W or k D o n e b y t h e G a s$

$C a l c u l a t in g C han g e in I n t er na l E n er g y$

$C a l c u l a t in g T o t a l H e a t S u ppl i e d$

$T h e W a y F or w a r d$

The Sigma Insight: First Law of Thermodynamics

Analyzing the Initial State

The Expansion

Force Equilibrium

The Master Equation

The Energy Split

Internal Energy

The First Law

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