An ideal monatomic gas of n moles is taken through a cycle WXYZW consisting of consecutive adiabatic

Q: Why are XY and ZW considered isobaric processes?

In a V-T diagram, the ideal gas law $PV = nRT$ can be rearranged as $V = (nR/P)T$. If the graph is a straight line passing through the origin, its slope $(nR/P)$ is constant, which means the pressure $P$ must be constant.

Q: Why did we use $C_P = 5R/2$?

The problem states it's an ideal monatomic gas. For monatomic gases, the degrees of freedom $f = 3$. Therefore, $C_V = fR/2 = 3R/2$, and $C_P = C_V + R = 5R/2$.

Q: Do we need to convert the volumes from $\text{cm}^3$ to $\text{m}^3$?

While it's always a good practice to work in SI units, in this specific calculation, the volume terms appear as a ratio $(V_W/V_X)^{5/3}$ and a difference $(V_Y - V_X)$ multiplied by $P_W$ which has a $1/V_W$ term. The conversion factor $10^{-6}$ perfectly cancels out in the numerator and denominator.

Q: How did we find $P_W$ without knowing the number of moles $n$?

We were given a crucial piece of information: $nRT_W = 1 \text{ J}$. According to the ideal gas law, $P_W V_W = nRT_W$. Therefore, $P_W V_W = 1 \text{ J}$, which allows us to find $P_W$ simply as $1/V_W$.

Q: Why is the work done in an isobaric process $P\Delta V$ related to heat?

From the first law of thermodynamics, $Q = \Delta U + W$. For an isobaric process, $W = P\Delta V = nR\Delta T$, and $\Delta U = nC_V\Delta T$. So $Q = n(C_V + R)\Delta T = nC_P\Delta T$. We used this directly to find the heat absorbed.

Question

An ideal monatomic gas of n moles is taken through a cycle WXYZW consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic V-T diagram. The volume of the gas at W, X and Y points are, $64 	ext{ cm}^3$, $125 	ext{ cm}^3$ and $250 	ext{ cm}^3$, respectively. If the absolute temperature of the gas $T_W$ at the point W is such that $nRT_W = 1 	ext{ J}$ (R is the universal gas constant), then the amount of heat absorbed (in J) by the gas along the path XY is ____

Sigma Percentile · Accepted Answer

### Decoding the V-T Graph: A Journey Through Thermodynamic Processes

Imagine you are a microscopic observer inside a container holding an ideal monatomic gas. The gas is undergoing a fascinating cyclic journey, expanding and compressing, heating up and cooling down. This journey is mapped out for us on a Volume-Temperature (V-T) diagram. Our mission? To find out exactly how much heat the gas absorbs during one specific leg of this journey: the path from state X to state Y.

### Analyzing the Setup

The first step in any thermodynamics problem is to decode the graph. We are given a cycle $WXYZW$ on a V-T diagram. The problem states it consists of consecutive adiabatic and isobaric processes. But which is which?

Look closely at the lines $XY$ and $ZW$. If you extend them, they pass straight through the origin. In a V-T diagram, a straight line passing through the origin means that Volume is directly proportional to Temperature ($V \propto T$). According to the ideal gas law, $PV = nRT$, or $V = \left(\frac{nR}{P}ight)T$. For the slope to be constant, the pressure $P$ must be constant. Therefore, **paths $XY$ and $ZW$ are isobaric processes**.

By elimination, the curved paths **$WX$ and $YZ$ must be the adiabatic processes**.

### The Master Equation

We need the heat absorbed during the isobaric expansion $XY$. The formula for heat exchanged at constant pressure is:

$Q_{XY} = n C_P \Delta T = n C_P (T_Y - T_X)$

For an ideal monatomic gas, the molar heat capacity at constant pressure is $C_P = \frac{5}{2}R$. Substituting this in, we get:

$Q_{XY} = n \left(\frac{5}{2}Right) (T_Y - T_X) = \frac{5}{2} (nRT_Y - nRT_X)$

Using the ideal gas law again, we can replace the temperature terms with pressure and volume:

$Q_{XY} = \frac{5}{2} (P_Y V_Y - P_X V_X)$

Since the process $XY$ is isobaric, the pressure remains constant, meaning $P_Y = P_X$. We can factor this out to get our master equation:

$Q_{XY} = \frac{5}{2} P_X (V_Y - V_X)$

We know the volumes $V_X$ and $V_Y$, but we are missing the pressure $P_X$.

### Bridging the Gap

To find $P_X$, we must look at the preceding process, the adiabatic path $WX$. The defining equation for an adiabatic process is $PV^\gamma = 	ext{constant}$. Therefore:

$P_X V_X^\gamma = P_W V_W^\gamma$

Rearranging this to solve for $P_X$, we get:

$P_X = P_W \left(\frac{V_W}{V_X}ight)^\gamma$

For a monatomic gas, the adiabatic index $\gamma = \frac{5}{3}$. But wait, we still don't know $P_W$!

Here is where the problem gives us a golden key: $nRT_W = 1 	ext{ J}$. From the ideal gas law, $P_W V_W = nRT_W$, which means $P_W V_W = 1 	ext{ J}$. This elegantly simplifies to:

$P_W = \frac{1}{V_W}$

Now, we substitute this back into our equation for $P_X$:

$P_X = \frac{1}{V_W} \left(\frac{V_W}{V_X}ight)^{\frac{5}{3}}$

### Final Calculation

We now have everything we need. Let's plug our expression for $P_X$ back into the master heat equation:

$Q_{XY} = \frac{5}{2} \left[ \frac{1}{V_W} \left(\frac{V_W}{V_X}ight)^{\frac{5}{3}} ight] (V_Y - V_X)$

Now, we substitute the given volume values: $V_W = 64 	ext{ cm}^3$, $V_X = 125 	ext{ cm}^3$, and $V_Y = 250 	ext{ cm}^3$. You might be worried about converting $	ext{cm}^3$ to $	ext{m}^3$. While it's a good habit, notice how the conversion factor $10^{-6}$ appears in the numerator $(V_Y - V_X)$ and the denominator $V_W$, and also inside the ratio $(V_W/V_X)$. It perfectly cancels out everywhere! We can safely use the raw numbers:

$Q_{XY} = \frac{5}{2} 	imes \frac{1}{64} 	imes \left(\frac{64}{125}ight)^{\frac{5}{3}} 	imes (250 - 125)$

Notice the beautiful numbers chosen by the examiner. $64$ is $4^3$ and $125$ is $5^3$. This makes applying the power of $\frac{5}{3}$ incredibly satisfying:

$Q_{XY} = \frac{5}{2} 	imes \frac{1}{64} 	imes \left(\frac{4^3}{5^3}ight)^{\frac{5}{3}} 	imes 125$

$Q_{XY} = \frac{5}{2} 	imes \frac{1}{64} 	imes \frac{4^5}{5^5} 	imes 125$

$Q_{XY} = \frac{5}{2} 	imes \frac{1}{64} 	imes \frac{1024}{3125} 	imes 125$

Simplifying the fractions:

$Q_{XY} = \frac{5}{2} 	imes 16 	imes \frac{1}{25} = \frac{80}{50} = 1.6 	ext{ J}$

And there we have it. Through careful analysis of the graph and methodical substitution, we found that the gas absorbs exactly **$1.60 	ext{ J}$** of heat during the isobaric expansion.

Enter Numerical Value:

Visualized Solution

$V - T Graph Analysis$

$Q_{X Y} Formula$

$Adiabatic Relation for P_{X}$

$Finding P_{W}$

$Substituting P_{W} into P_{X}$

$Master Equation for Q_{X Y}$

$Substituting Volume Values$

$Simplifying the Powers$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Thermodynamic Processes

Decoding the V-T Graph

Analyzing the Setup

The Master Equation

Bridging the Gap

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]

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)

One mole of a monatomic ideal gas undergoes the cyclic process J $\to$ K $\to$ L $\to$ M $\to$ J, as shown in the P-T diagram. Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [R is the gas constant.]

List-I

List-II

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V−T Graph Analysis

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Adiabatic Relation for PX​

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Master Equation for QXY​

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

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

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

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