An ideal monoatomic gas at temperature `27^@C` and pressure `10^6N//m^2` occupies 10L volume. 10,000 cal of heat is added to the system without changing the volume. Calculate the change in temperature of the gas. Given : `R=8.31 J//mol-K` and `J=4.18J//cal.`

Explanation

To solve the problem step by step, we will follow the outlined procedure:

### Step 1: Convert the initial temperature to Kelvin
The initial temperature is given as \(27^\circ C\). To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
So,
\[ T = 27 + 273.15 = 300.15 \, K \approx 300 \, K \]

### Step 2: Convert the volume from liters to cubic meters
The volume is given as \(10 \, L\). To convert liters to cubic meters, we use the conversion:
\[ 1 \, L = 10^{-3} \, m^3 \]
Thus,
\[ V = 10 \, L = 10 \times 10^{-3} \, m^3 = 0.01 \, m^3 \]

### Step 3: Use the ideal gas equation to find the number of moles
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \(P = 10^6 \, N/m^2\)
- \(V = 0.01 \, m^3\)
- \(R = 8.31 \, J/(mol \cdot K)\)
- \(T = 300 \, K\)

Rearranging the equation to solve for \(n\):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{(10^6 \, N/m^2)(0.01 \, m^3)}{(8.31 \, J/(mol \cdot K))(300 \, K)} \]
Calculating:
\[ n = \frac{10000}{2493} \approx 4.01 \, mol \approx 4 \, mol \]

### Step 4: Calculate the molar heat capacity at constant volume (\(C_v\))
For a monoatomic ideal gas, the molar heat capacity at constant volume is given by:
\[ C_v = \frac{3}{2} R \]
Substituting \(R\):
\[ C_v = \frac{3}{2} \times 8.31 \, J/(mol \cdot K) = 12.465 \, J/(mol \cdot K) \]
To convert \(C_v\) to calories, we use the conversion \(1 \, cal = 4.18 \, J\):
\[ C_v = \frac{12.465}{4.18} \approx 2.98 \, cal/(mol \cdot K) \approx 3 \, cal/(mol \cdot K) \]

### Step 5: Apply the first law of thermodynamics
Since the volume is constant, the work done \(W = 0\). Thus, from the first law of thermodynamics:
\[ Q = \Delta U \]
Where \(\Delta U = nC_v \Delta T\). Rearranging gives:
\[ \Delta T = \frac{Q}{nC_v} \]
Substituting the values:
\[ Q = 10000 \, cal, \quad n = 4 \, mol, \quad C_v = 3 \, cal/(mol \cdot K) \]
Calculating:
\[ \Delta T = \frac{10000}{4 \times 3} = \frac{10000}{12} \approx 833.33 \, K \]

### Final Answer
The change in temperature of the gas is approximately \(833.33 \, K\).

---