70 calories of heat required to raise the temperature of 2 moles of an ideal gas at constant pressure from `30^@Cto 35^@C.` The amount of heat required (in calories) to raise the temperature of the same gas through the same range `(30^@C to 35^@C)` at constant volume is:

Answer

To solve the problem of finding the amount of heat required to raise the temperature of 2 moles of an ideal gas from \(30^\circ C\) to \(35^\circ C\) at constant volume, we can follow these steps:

### Step 1: Identify the Given Data
- Heat required at constant pressure (\(q_p\)) = 70 calories
- Number of moles (\(n\)) = 2 moles
- Temperature change (\(\Delta T\)) = \(35^\circ C - 30^\circ C = 5^\circ C\)

### Step 2: Use the Formula for Heat at Constant Pressure
The formula for heat at constant pressure is given by:
\[ q_p = n C_p \Delta T \]
We can rearrange this to find \(C_p\):
\[ C_p = \frac{q_p}{n \Delta T} \]

### Step 3: Calculate \(C_p\)
Substituting the known values into the equation:
\[ C_p = \frac{70 \text{ calories}}{2 \text{ moles} \times 5 \text{ °C}} = \frac{70}{10} = 7 \text{ calories/(mole·°C)} \]

### Step 4: Relate \(C_p\) and \(C_v\)
For an ideal gas, the relationship between \(C_p\) and \(C_v\) is given by:
\[ C_p - C_v = R \]
Where \(R\) is the gas constant. In calories, \(R = 2 \text{ calories/(mole·°C)}\).

### Step 5: Calculate \(C_v\)
Using the relationship:
\[ C_v = C_p - R \]
Substituting the values:
\[ C_v = 7 \text{ calories/(mole·°C)} - 2 \text{ calories/(mole·°C)} = 5 \text{ calories/(mole·°C)} \]

### Step 6: Calculate Heat at Constant Volume
The formula for heat at constant volume is:
\[ q_v = n C_v \Delta T \]
Substituting the known values:
\[ q_v = 2 \text{ moles} \times 5 \text{ calories/(mole·°C)} \times 5 \text{ °C} = 50 \text{ calories} \]

### Final Answer
The amount of heat required to raise the temperature of the gas from \(30^\circ C\) to \(35^\circ C\) at constant volume is **50 calories**.

---