70 calories of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from `40^(@)C` to `45^(@)C` (R=2cal/mol`-.^(@)C`). The gas may be

Options

A. `H_(2)`
B. He
C. `CO_(2)`
D. `NH_(3)`

Answer

To solve the problem, we need to determine the type of gas based on the heat required to raise its temperature at constant pressure. Here’s a step-by-step solution:

### Step 1: Understand the given data
We are given:
- Heat required (ΔQ) = 70 calories
- Number of moles (n) = 2 moles
- Initial temperature (T1) = 40°C
- Final temperature (T2) = 45°C
- R (gas constant) = 2 cal/mol·°C

### Step 2: Calculate the change in temperature (ΔT)
The change in temperature (ΔT) can be calculated as:
\[ \Delta T = T2 - T1 = 45°C - 40°C = 5°C \]

### Step 3: Use the formula for heat at constant pressure
The heat added at constant pressure is given by the formula:
\[ \Delta Q = n C_p \Delta T \]
Where \(C_p\) is the molar heat capacity at constant pressure.

### Step 4: Rearranging the formula
We can rearrange the formula to find \(C_p\):
\[ C_p = \frac{\Delta Q}{n \Delta T} \]

### Step 5: Substitute the known values
Substituting the known values into the equation:
\[ C_p = \frac{70 \text{ cal}}{2 \text{ moles} \times 5°C} = \frac{70}{10} = 7 \text{ cal/mol·°C} \]

### Step 6: Relate \(C_p\) to \(\gamma\)
We know that:
\[ C_p = \frac{\gamma R}{\gamma - 1} \]
Substituting \(R = 2 \text{ cal/mol·°C}\) into the equation gives us:
\[ 7 = \frac{\gamma \cdot 2}{\gamma - 1} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ 7(\gamma - 1) = 2\gamma \]
Expanding this:
\[ 7\gamma - 7 = 2\gamma \]
Rearranging terms:
\[ 7\gamma - 2\gamma = 7 \]
\[ 5\gamma = 7 \]
\[ \gamma = \frac{7}{5} \]

### Step 8: Identify the type of gas
The value of \(\gamma = \frac{7}{5}\) corresponds to a diatomic gas. Among the options provided, we can identify that:
- Helium is monatomic.
- CO2 is not diatomic.
- NH3 is not diatomic.
- H2 is diatomic.

Thus, the gas must be **H2** (hydrogen gas).

### Final Answer
The gas may be **H2** (hydrogen gas).

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