To find the amount of heat taken by 5 moles of oxygen when heated at constant volume from \(10^\circ C\) to \(20^\circ C\), we can follow these steps:
### Step 1: Understand the relationship between heat, internal energy, and work
At constant volume, the heat added to the system is equal to the change in internal energy:
\[
Q = \Delta U
\]
Since there is no work done at constant volume (\(W = 0\)), we have:
\[
Q = \Delta U
\]
### Step 2: Use the formula for change in internal energy
The change in internal energy (\(\Delta U\)) can be expressed in terms of the number of moles (\(n\)) and the molar specific heat at constant volume (\(C_v\)):
\[
\Delta U = n C_v \Delta T
\]
where \(\Delta T\) is the change in temperature.
### Step 3: Calculate \(\Delta T\)
Given that the initial temperature is \(10^\circ C\) and the final temperature is \(20^\circ C\):
\[
\Delta T = 20 - 10 = 10^\circ C
\]
### Step 4: Find \(C_v\) using the relationship with \(C_p\)
We know that:
\[
C_v = C_p - R
\]
Given:
- \(C_p = 7.03 \, \text{cal/mol} \cdot ^\circ C\)
- \(R = 8.31 \, \text{J/mol} \cdot ^\circ C\)
First, we need to convert \(R\) from Joules to calories. Using the conversion \(1 \, \text{cal} = 4.184 \, \text{J}\):
\[
R = \frac{8.31 \, \text{J/mol} \cdot ^\circ C}{4.184 \, \text{J/cal}} \approx 1.987 \, \text{cal/mol} \cdot ^\circ C
\]
Now we can calculate \(C_v\):
\[
C_v = C_p - R = 7.03 \, \text{cal/mol} \cdot ^\circ C - 1.987 \, \text{cal/mol} \cdot ^\circ C \approx 5.043 \, \text{cal/mol} \cdot ^\circ C
\]
### Step 5: Substitute values into the internal energy equation
Now we can substitute \(n = 5 \, \text{mol}\), \(C_v \approx 5.043 \, \text{cal/mol} \cdot ^\circ C\), and \(\Delta T = 10^\circ C\) into the equation for \(\Delta U\):
\[
\Delta U = 5 \, \text{mol} \times 5.043 \, \text{cal/mol} \cdot ^\circ C \times 10^\circ C
\]
\[
\Delta U \approx 5 \times 5.043 \times 10 \approx 252.15 \, \text{cal}
\]
### Step 6: Final result
The amount of heat taken by 5 moles of oxygen when heated at constant volume from \(10^\circ C\) to \(20^\circ C\) is approximately:
\[
Q \approx 252 \, \text{cal}
\]
