A monatomic ideal gas is heated at constant volume until its pressure is doubled. It is again heated at constant pressure, until its volume is doubled. Find molar specific heat for the whole process.

Options

A. `(3R)/(2)`
B. `(5R)/(2)`
C. `(13R)/(6)`
D. `(7R)/(2)`

Answer

To find the molar specific heat for the entire process involving a monatomic ideal gas, we will analyze the two stages of heating:

1. Heating at constant volume until the pressure is doubled.
2. Heating at constant pressure until the volume is doubled.

### Step 1: Analyze the First Process (Constant Volume Heating)

In the first process, we start with an initial state (1) where the pressure is \( P_1 \) and the volume is \( V_1 \). The gas is heated at constant volume until its pressure doubles, reaching state (2) with pressure \( P_2 = 2P_1 \).

Using the ideal gas law \( PV = nRT \), we can relate the temperatures:
- At state 1: \( P_1 V_1 = nRT_1 \)
- At state 2: \( P_2 V_1 = nRT_2 \)

Since the volume is constant, we can write:
\[ \frac{P_2}{P_1} = \frac{T_2}{T_1} \]
Substituting \( P_2 = 2P_1 \):
\[ \frac{2P_1}{P_1} = \frac{T_2}{T_1} \implies T_2 = 2T_1 \]

### Step 2: Analyze the Second Process (Constant Pressure Heating)

In the second process, we start from state (2) and heat the gas at constant pressure until the volume doubles, reaching state (3) with volume \( V_3 = 2V_1 \).

Using the ideal gas law again:
- At state 2: \( P_2 V_2 = nRT_2 \)
- At state 3: \( P_2 V_3 = nRT_3 \)

Since pressure is constant, we can relate the temperatures:
\[ \frac{T_3}{T_2} = \frac{V_3}{V_2} \]
Substituting \( V_3 = 2V_1 \) and \( V_2 = V_1 \):
\[ \frac{T_3}{T_2} = \frac{2V_1}{V_1} \implies T_3 = 2T_2 \]
Substituting \( T_2 = 2T_1 \):
\[ T_3 = 2(2T_1) = 4T_1 \]

### Step 3: Calculate the Heat Added in Each Process

1. **Heat added in the first process (constant volume)**:
\[ Q_1 = nC_v(T_2 - T_1) = nC_v(2T_1 - T_1) = nC_vT_1 \]
For a monatomic ideal gas, \( C_v = \frac{3}{2}R \):
\[ Q_1 = n \left(\frac{3}{2}R\right) T_1 \]

2. **Heat added in the second process (constant pressure)**:
\[ Q_2 = nC_p(T_3 - T_2) = nC_p(4T_1 - 2T_1) = nC_p(2T_1) \]
For a monatomic ideal gas, \( C_p = \frac{5}{2}R \):
\[ Q_2 = n \left(\frac{5}{2}R\right)(2T_1) = n(5R)T_1 \]

### Step 4: Total Heat Added

The total heat added during the entire process is:
\[ Q = Q_1 + Q_2 = n \left(\frac{3}{2}R T_1 + 5RT_1\right) = n \left(\frac{3}{2}R + 5R\right) T_1 = n \left(\frac{3 + 10}{2}R\right) T_1 = n \left(\frac{13}{2}R\right) T_1 \]

### Step 5: Calculate the Molar Specific Heat

The molar specific heat \( C \) for the entire process is given by:
\[ C = \frac{Q}{n(T_3 - T_1)} = \frac{n \left(\frac{13}{2}R\right) T_1}{n(4T_1 - T_1)} = \frac{\frac{13}{2}R T_1}{3T_1} = \frac{13}{6}R \]

### Final Answer

Thus, the molar specific heat for the whole process is:
\[ C = \frac{13R}{6} \]