600J of heat is added to a monoatomic gas in a process in which the gas performs a work of 150J. The molar heat capacity for the process is

Options

A. (a) 3R
B. (b) 4R
C. (c) 2R
D. (d) 6R

Explanation

To find the molar heat capacity for the process, we will use the first law of thermodynamics and the properties of a monoatomic gas.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Heat added to the gas (Q) = 600 J
- Work done by the gas (W) = 150 J

2. **Apply the First Law of Thermodynamics:**
The first law of thermodynamics states:
\[ \Delta U = Q - W \]
where \(\Delta U\) is the change in internal energy.

3. **Calculate the Change in Internal Energy (\(\Delta U\)):**
Substitute the values of Q and W into the equation:
\[ \Delta U = 600 J - 150 J = 450 J \]

4. **Relate Change in Internal Energy to Molar Heat Capacity:**
For a monoatomic gas, the change in internal energy can also be expressed as:
\[ \Delta U = n C_v \Delta T \]
where \(C_v\) is the molar heat capacity at constant volume.

5. **Substituting for \(C_v\):**
For a monoatomic gas, the molar heat capacity at constant volume is:
\[ C_v = \frac{3}{2} R \]
Thus, we can write:
\[ 450 J = n \left(\frac{3}{2} R\right) \Delta T \]

6. **Express \(\Delta T\) in terms of \(n\) and \(C\):**
The heat added can also be expressed as:
\[ Q = n C \Delta T \]
where \(C\) is the molar heat capacity for the process.

7. **Rearranging for Molar Heat Capacity (C):**
From the equation \(Q = n C \Delta T\), we can express \(C\) as:
\[ C = \frac{Q}{n \Delta T} \]

8. **Substituting \(\Delta T\) from the Internal Energy Equation:**
From the previous step, we have:
\[ n \Delta T = \frac{450 J}{\frac{3}{2} R} \]
Therefore, substituting this into the equation for \(C\):
\[ C = \frac{600 J}{\frac{450 J}{\frac{3}{2} R}} = \frac{600 J \cdot \frac{3}{2} R}{450 J} \]

9. **Simplifying the Expression:**
\[ C = \frac{600 \cdot 3R}{450 \cdot 2} = \frac{1800R}{900} = 2R \]

10. **Final Answer:**
The molar heat capacity for the process is:
\[ C = 2R \]