One mole of an ideal monoatomic gas at `27^(@)C` undergoes the process in which `TalphaV^(3).` Then calculate the heat absorbed (in calories) when gas doubles it's volume `(R=2"cal//mol"-K)`

Answer

To solve the problem step by step, we will follow the provided information and apply the relevant thermodynamic principles.

### Step 1: Understand the Given Information
We have one mole of an ideal monoatomic gas at an initial temperature of \(27^\circ C\) (which is \(300 K\)). The gas undergoes a process where the temperature \(T\) is proportional to the volume \(V\) cubed, i.e., \(T \propto V^3\). We need to calculate the heat absorbed when the volume of the gas doubles.

### Step 2: Determine Initial and Final Temperatures
Since the volume doubles, we can express this mathematically:
- Let the initial volume be \(V_1\).
- The final volume will be \(V_2 = 2V_1\).

From the relationship \(T \propto V^3\):
\[ \frac{T_2}{T_1} = \left(\frac{V_2}{V_1}\right)^3 = (2)^3 = 8 \]
Thus, we can find \(T_2\):
\[ T_2 = 8 \times T_1 = 8 \times 300 K = 2400 K \]

### Step 3: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
For our case, we can express the temperature in terms of pressure and volume:
\[ T = \frac{PV}{nR} \]
Since \(n = 1\) mole and \(R = 2 \, \text{cal/mol-K}\), we can write:
\[ T \propto V^3 \implies PV \propto V^3 \implies PV^{n} = \text{constant} \]
Here, we need to find \(n\). From the relation \(PV^n = \text{constant}\), we can deduce that:
\[ n = -2 \]

### Step 4: Calculate the Heat Capacity
For a monoatomic ideal gas, the heat capacities are:
\[ C_V = \frac{3}{2}R \quad \text{and} \quad C_P = \frac{5}{2}R \]
The effective heat capacity \(C\) for the process can be calculated using:
\[ C = C_V + \frac{R}{1-n} \]
Substituting \(C_V\) and \(n\):
\[ C = \frac{3}{2}R + \frac{R}{1 - (-2)} = \frac{3}{2}R + \frac{R}{3} \]
Finding a common denominator (which is 6):
\[ C = \frac{9R}{6} + \frac{2R}{6} = \frac{11R}{6} \]

### Step 5: Calculate the Heat Absorbed
Using the formula for heat absorbed:
\[ Q = nC\Delta T \]
Where:
- \(n = 1\) mole
- \(C = \frac{11R}{6}\)
- \(\Delta T = T_2 - T_1 = 2400 K - 300 K = 2100 K\)

Substituting the values:
\[ Q = 1 \times \frac{11R}{6} \times 2100 \]
Substituting \(R = 2 \, \text{cal/mol-K}\):
\[ Q = \frac{11 \times 2}{6} \times 2100 = \frac{22}{6} \times 2100 = \frac{22 \times 2100}{6} = 7700 \, \text{calories} \]

### Final Answer
The heat absorbed when the gas doubles its volume is \(Q = 7700 \, \text{calories}\).

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