Neon is a monatomic gas with a molar heat capacity at constant volume of 12.66 J/(mol K). Two moles of neon gas enclosed in a constant volume system receive 5250 J of heat. If the gas was initially at 293 K, what is the final temperature of the neon?

Options

A. 200 K
B. 300 K
C. 400 K
D. 500 K

Answer

To solve the problem step by step, we will use the formula for heat transfer at constant volume and the given data.

### Step 1: Identify the given data
- Molar heat capacity at constant volume (C_v) = 12.66 J/(mol K)
- Number of moles (n) = 2 moles
- Heat added (Q) = 5250 J
- Initial temperature (T_initial) = 293 K

### Step 2: Use the formula for heat transfer at constant volume
The formula for heat transfer at constant volume is given by:
\[ Q = n C_v \Delta T \]
where \(\Delta T\) is the change in temperature.

### Step 3: Rearrange the formula to find \(\Delta T\)
We can rearrange the formula to solve for \(\Delta T\):
\[ \Delta T = \frac{Q}{n C_v} \]

### Step 4: Substitute the values into the equation
Now, we can substitute the values we have into the equation:
\[ \Delta T = \frac{5250 \, \text{J}}{2 \, \text{moles} \times 12.66 \, \text{J/(mol K)}} \]

### Step 5: Calculate \(\Delta T\)
Calculating the denominator:
\[ 2 \times 12.66 = 25.32 \, \text{J/K} \]

Now substituting this back into the equation for \(\Delta T\):
\[ \Delta T = \frac{5250}{25.32} \approx 207.1 \, \text{K} \]

### Step 6: Calculate the final temperature
Now that we have \(\Delta T\), we can find the final temperature:
\[ T_{final} = T_{initial} + \Delta T \]
\[ T_{final} = 293 \, \text{K} + 207.1 \, \text{K} \]
\[ T_{final} \approx 500.1 \, \text{K} \]

### Conclusion
The final temperature of the neon gas is approximately **500.1 K**.

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