Two moles of an ideal gas at temperature `T_(0) = 300 K` was cooled isochorically so that the pressure was reduced to half. Then, in an isobaric process, the gas expanded till its temperature got back to the initial value. Find the total amount of heat absorbed by the gas in the processs

Options

A. `150 R J`
B. `300 R J`
C. `75 R J`
D. `100 R J`

Explanation

To solve the problem step by step, we will analyze the two processes the gas undergoes: the isochoric cooling and the isobaric expansion.

### Step 1: Initial Conditions
We start with 2 moles of an ideal gas at an initial temperature \( T_0 = 300 \, K \).

### Step 2: Isochoric Cooling
In the first process, the gas is cooled isochorically (constant volume) until the pressure is halved.

- Let the initial pressure be \( P_0 \).
- The final pressure after cooling is \( P_f = \frac{P_0}{2} \).

Using the ideal gas law, we can relate the pressures and temperatures:
\[ \frac{P_0}{T_0} = \frac{P_f}{T_f} \]
Substituting the known values:
\[ \frac{P_0}{300} = \frac{\frac{P_0}{2}}{T_f} \]
Cross-multiplying gives:
\[ P_0 T_f = \frac{P_0}{2} \cdot 300 \]
Cancelling \( P_0 \) (assuming \( P_0 \neq 0 \)):
\[ T_f = \frac{300}{2} = 150 \, K \]

### Step 3: Change in Internal Energy (Isochoric Process)
The change in internal energy \( \Delta U_1 \) for an ideal gas during an isochoric process is given by:
\[ \Delta U_1 = n C_v \Delta T \]
Where \( \Delta T = T_f - T_0 = 150 - 300 = -150 \, K \).

Substituting the values:
\[ \Delta U_1 = 2 C_v (-150) = -300 C_v \]

Since the process is isochoric, the heat absorbed \( Q_1 \) is equal to the change in internal energy:
\[ Q_1 = \Delta U_1 = -300 C_v \]

### Step 4: Isobaric Expansion
In the second process, the gas expands isobarically (constant pressure) back to the initial temperature \( T_0 = 300 \, K \).

The change in enthalpy \( \Delta H \) for an ideal gas during an isobaric process is given by:
\[ \Delta H = n C_p \Delta T \]
Where \( \Delta T = T_0 - T_f = 300 - 150 = 150 \, K \).

Substituting the values:
\[ \Delta H = 2 C_p (150) = 300 C_p \]

Since the process is isobaric, the heat absorbed \( Q_2 \) is equal to the change in enthalpy:
\[ Q_2 = \Delta H = 300 C_p \]

### Step 5: Total Heat Absorbed
The total heat absorbed by the gas in the entire process is:
\[ Q_{\text{total}} = Q_1 + Q_2 = -300 C_v + 300 C_p \]

Using the relation \( C_p - C_v = R \):
\[ Q_{\text{total}} = 300 (C_p - C_v) = 300 R \]

### Final Answer
Thus, the total amount of heat absorbed by the gas in the process is:
\[ Q_{\text{total}} = 300 R \, \text{Joules} \]