An ideal gas has a specific heat at constant pressure Cp = 5R/2. The gas is kept in a closed vessel

Q: Why is the work done by the gas zero?

Work done by a gas is given by $W = p\Delta V$. Since the gas is kept in a closed, rigid vessel, its volume cannot change ($\Delta V = 0$). Therefore, the gas does no work on its surroundings.

Q: Why do we use $C_V$ when the problem gives $C_p$?

The process of heating the gas happens in a closed vessel, meaning it's an isochoric (constant volume) process. The heat exchanged in such a process is governed by $C_V$ ($Q = nC_V\Delta T$). We use the given $C_p$ along with Mayer's relation ($C_p - C_V = R$) to find the required $C_V$.

Q: What happened to the universal gas constant $R$ in the calculation?

When we substitute the number of moles $n = \frac{p_1 V}{R T_1}$ and $C_V = \frac{3R}{2}$ into the heat equation $Q = n C_V \Delta T$, the $R$ in the denominator of $n$ cancels out with the $R$ in the numerator of $C_V$. This elegant cancellation saves us from plugging in the numerical value of $R$!

Q: Can we find the final pressure without calculating the number of moles?

Absolutely! Since the volume is constant, you can directly use Gay-Lussac's Law, which states that pressure is directly proportional to absolute temperature ($p_1/T_1 = p_2/T_2$). Once you have the final temperature $T_2$, finding $p_2$ is just a simple ratio calculation.

Q: What type of gas is this based on the given specific heat?

We are given $C_p = \frac{5R}{2}$. Using Mayer's relation, $C_V = \frac{3R}{2}$. The specific heat at constant volume is related to the degree of freedom $f$ by $C_V = \frac{f}{2}R$. This gives $f = 3$, which tells us that the gas is monatomic (like Helium or Argon).

Question

An ideal gas has a specific heat at constant pressure $C_p = \frac{5R}{2}$. The gas is kept in a closed vessel of volume $0.0083 	ext{ m}^3$, at a temperature of $300 	ext{ K}$ and a pressure of $1.6 	imes 10^6 	ext{ N/m}^2$. An amount of $2.49 	imes 10^4 	ext{ J}$ of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas.

Sigma Percentile · Accepted Answer

### Setting the Stage: The Closed Vessel

Imagine a closed, rigid vessel containing an ideal gas. The walls are fixed, meaning the volume cannot change. We are given the initial state of the gas: its volume $V = 0.0083 	ext{ m}^3$, temperature $T_1 = 300 	ext{ K}$, and pressure $p_1 = 1.6 	imes 10^6 	ext{ N/m}^2$. We are also given its specific heat at constant pressure, $C_p = \frac{5R}{2}$.

### The First Law: Where Does the Heat Go?

Now, we supply an amount of heat energy $Q = 2.49 	imes 10^4 	ext{ J}$ to this gas. According to the first law of thermodynamics, the heat supplied goes into doing work and changing the internal energy:

$$Q = \Delta U + W$$

But wait, the vessel is closed! What does that mean for the work done? Since the volume is constant ($\Delta V = 0$), the work done by the gas is zero ($W = 0$). This means all the heat supplied goes entirely into increasing the internal energy of the gas.

### Unlocking the Specific Heat

We know that the change in internal energy is given by:

$$Q = \Delta U = n C_V \Delta T$$

We need the specific heat at constant volume, $C_V$. We are given $C_p$. Remember Mayer's relation? Let's use it to find $C_V$:

$$C_p - C_V = R$$

$$C_V = C_p - R = \frac{5R}{2} - R = \frac{3R}{2}$$

### The Ideal Gas Law to the Rescue

We also need the number of moles, $n$. We can find this using the ideal gas equation for the initial state:

$$p_1 V = n R T_1 \implies n = \frac{p_1 V}{R T_1}$$

Let's substitute $n$ and $C_V$ back into our heat equation. We get:

$$Q = \left(\frac{p_1 V}{R T_1}ight) \left(\frac{3R}{2}ight) \Delta T$$

Notice how the universal gas constant $R$ beautifully cancels out! This makes our calculation much simpler.

### The Final Calculation: Temperature and Pressure

Rearranging for $\Delta T$, we get:

$$\Delta T = \frac{2 Q T_1}{3 p_1 V}$$

Now, let's plug in the given values to find the change in temperature:

$$\Delta T = \frac{2 	imes (2.49 	imes 10^4) 	imes 300}{3 	imes (1.6 	imes 10^6) 	imes 0.0083} = 375 	ext{ K}$$

The final temperature $T_2$ is simply the initial temperature plus this change:

$$T_2 = T_1 + \Delta T = 300 + 375 = 675 	ext{ K}$$

Next, we need the final pressure. Since the volume is constant, Gay-Lussac's law tells us that pressure is directly proportional to temperature:

$$\frac{p_2}{p_1} = \frac{T_2}{T_1} \implies p_2 = p_1 \left( \frac{T_2}{T_1} ight)$$

Let's substitute the values:

$$p_2 = (1.6 	imes 10^6) 	imes \left( \frac{675}{300} ight) = 3.6 	imes 10^6 	ext{ N/m}^2$$

And there we have it! The final temperature is **$675 	ext{ K}$** and the final pressure is **$3.6 	imes 10^6 	ext{ N/m}^2$**.

Visualized Solution

$V = constant$

$First Law of Thermodynamics$

$Internal Energy Change$

$Mayer’s Relation$

$Ideal Gas Law$

$Substituting Values$

$Calculating Δ T$

$Final Temperature$

$Gay-Lussac’s Law$

$Final Pressure$

$Conclusion$

The Sigma Insight: First Law of Thermodynamics

Setting the Stage

The First Law

Unlocking the Specific Heat

The Ideal Gas Law to the Rescue

The Final Calculation

Similar Questions

An ideal gas expands in volume from $1 im es 1 0^{- 3} m^{3}$ to $1 im es 1 0^{- 2} m^{3}$ at $300 K$ against a constant pressure of $1 im es 1 0^{5} Nm^{- 2}$ . The work done is

A piston filled with $0.04$ mole of an ideal gas expands reversibly from $50.0 mL$ to $375 mL$ at a constant temperature of $37. 0^{\circ} C$ . As it does so, it absorbs $208 J$ of heat. The values of $q$ and $W$ for the process will be ( $R = 8.314 J / mol K$ , $ln 7.5 = 2.01$ )

A thermally insulated vessel contains an ideal gas of molecular mass $M$ and ratio of specific heats $γ$ . It is moving with speed $v$ and its suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by

5 moles of an ideal gas at $100 K$ are allowed to undergo reversible compression till its temperature becomes $200 K$ . If $C_{V} = 28 JK^{- 1} mol^{- 1}$ , calculate $Δ U$ and $Δ (p V)$ for this process. ( $R = 8.0 JK^{- 1} mol^{- 1}$ )

Visualized Solution

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

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

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Gay-Lussac’s Law

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

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