One mole of Argon is heated using `PV^(5//2)` = constant . By what amount heat is absorded during the
'process , when temperture changes by `DeltaT=26 K`.

Options

A. `100J`
B. `200J`
C. `180 J`
D. `208 J`

Explanation

To solve the problem of how much heat is absorbed when one mole of Argon is heated under the condition \(PV^{\frac{5}{2}} = \text{constant}\) with a temperature change of \(\Delta T = 26 \, K\), we will follow these steps:

### Step 1: Identify the type of process
The equation \(PV^{\frac{5}{2}} = \text{constant}\) indicates that this is a polytropic process. The value of \(n\) in the polytropic equation \(PV^n = \text{constant}\) is \(\frac{5}{2}\).

### Step 2: Determine the heat capacity for the process
For a polytropic process, the heat capacity \(C\) can be calculated using the formula:
\[ C = R \left( \gamma - 1 \right) + R \left( 1 - n \right) \]
where \(R\) is the universal gas constant, \(\gamma\) is the adiabatic index (ratio of specific heats), and \(n\) is the polytropic index.

### Step 3: Find the value of \(\gamma\) for Argon
Argon is a monoatomic gas, and for monoatomic gases, \(\gamma\) is given by:
\[ \gamma = \frac{C_p}{C_v} = \frac{5}{3} \]

### Step 4: Substitute the values into the heat capacity formula
Substituting \(\gamma\) and \(n\) into the heat capacity formula:
\[ C = R \left( \frac{5}{3} - 1 \right) + R \left( 1 - \frac{5}{2} \right) \]
Calculating this:
\[ C = R \left( \frac{2}{3} \right) + R \left( -\frac{3}{2} \right) = R \left( \frac{2}{3} - \frac{3}{2} \right) \]
Finding a common denominator (which is 6):
\[ C = R \left( \frac{4}{6} - \frac{9}{6} \right) = R \left( -\frac{5}{6} \right) \]

### Step 5: Calculate the heat absorbed
The heat absorbed \(Q\) during the process can be calculated using the formula:
\[ Q = nC\Delta T \]
Given that \(n = 1\) mole, \(\Delta T = 26 \, K\), and substituting \(C = \frac{5}{6}R\):
\[ Q = 1 \cdot \frac{5}{6}R \cdot 26 \]
Substituting the value of \(R = 8.314 \, J/(mol \cdot K)\):
\[ Q = \frac{5}{6} \cdot 8.314 \cdot 26 \]
Calculating this:
\[ Q = \frac{5 \cdot 8.314 \cdot 26}{6} \approx 180 \, J \]

### Final Answer
The amount of heat absorbed during the process is approximately \(180 \, J\).

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