What amount of heat must be supplied to 35 g of oxygen at room temperature to raise its temperature by `80^(@)C` at constant volume (molecular mass of oxygen is 32 and R= 8.3 j `mol^(-1) K^(-1))` a) 1.52 KJ b) 3.23 KJ c) 1.81 KJ d) 1.62 KJ

Options

A. 1.52 KJ
B. 3.23 KJ
C. 1.81 KJ
D. 1.62 KJ

Explanation

To find the amount of heat that must be supplied to 35 g of oxygen to raise its temperature by \(80^\circ C\) at constant volume, we can follow these steps:

### Step 1: Calculate the number of moles of oxygen
The molecular mass of oxygen (O\(_2\)) is given as 32 g/mol. We can calculate the number of moles (n) using the formula:
\[ n = \frac{\text{mass}}{\text{molecular mass}} \]
Substituting the values:
\[ n = \frac{35 \, \text{g}}{32 \, \text{g/mol}} = 1.09375 \, \text{mol} \]

### Step 2: Determine the specific heat capacity at constant volume (C\(_V\))
For diatomic gases like oxygen, the relationship between the specific heat capacities at constant pressure (C\(_P\)) and constant volume (C\(_V\)) is given by:
\[ C_P - C_V = R \]
Where R is the gas constant, \(R = 8.3 \, \text{J/(mol K)}\).

For diatomic gases, the ratio \(\frac{C_P}{C_V}\) is typically \(\frac{7}{5}\). Thus, we can express \(C_P\) in terms of \(C_V\):
\[ C_P = \frac{7}{5} C_V \]
Substituting this into the first equation:
\[ \frac{7}{5} C_V - C_V = R \]
This simplifies to:
\[ \frac{2}{5} C_V = R \]
Now, solving for \(C_V\):
\[ C_V = \frac{5R}{2} = \frac{5 \times 8.3}{2} = 20.75 \, \text{J/(mol K)} \]

### Step 3: Calculate the amount of heat (Q) required
The heat added at constant volume can be calculated using the formula:
\[ Q = n C_V \Delta T \]
Where \(\Delta T\) is the change in temperature. Here, \(\Delta T = 80 \, \text{K}\) (since the change in Celsius is equivalent to the change in Kelvin).

Substituting the values:
\[ Q = 1.09375 \, \text{mol} \times 20.75 \, \text{J/(mol K)} \times 80 \, \text{K} \]
Calculating this:
\[ Q = 1.09375 \times 20.75 \times 80 = 1811.25 \, \text{J} \]
Converting to kilojoules:
\[ Q = \frac{1811.25}{1000} = 1.81125 \, \text{kJ} \approx 1.81 \, \text{kJ} \]

### Final Answer
The amount of heat required is approximately **1.81 kJ**.