Two moles of a certain ideal gas at 300K is cooled at constant volume so that the pressure is reduced to half the original value. Now the gas is heated at constant pressure so that its temperature becomes equal to the initial temperature. Find the total amount of heat absorbed by the gas in the process.

Answer

To solve the problem step by step, we will analyze the two processes the gas undergoes: cooling at constant volume and heating at constant pressure.

### Step 1: Analyze the Initial Conditions
We have 2 moles of an ideal gas at an initial temperature \( T_i = 300 \, K \).

### Step 2: Cooling at Constant Volume
During the first process, the gas is cooled at constant volume, and the pressure is reduced to half the original value.

Using the ideal gas law:
\[ PV = nRT \]
At constant volume, we can express the relationship between pressure and temperature:
\[ \frac{P_i}{T_i} = \frac{P_f}{T_f} \]
Where:
- \( P_i \) is the initial pressure
- \( P_f = \frac{P_i}{2} \) is the final pressure
- \( T_i = 300 \, K \) is the initial temperature
- \( T_f \) is the final temperature after cooling

Substituting the known values:
\[ \frac{P_i}{300} = \frac{\frac{P_i}{2}}{T_f} \]
Cross-multiplying gives:
\[ P_i T_f = \frac{P_i}{2} \cdot 300 \]
Cancelling \( P_i \) from both sides (assuming \( P_i \neq 0 \)):
\[ T_f = \frac{300}{2} = 150 \, K \]

### Step 3: Calculate Heat Released During Cooling
Since the cooling process is at constant volume, the heat \( Q_1 \) released can be calculated using:
\[ Q_1 = n C_v (T_f - T_i) \]
Where \( C_v \) is the molar heat capacity at constant volume. The change in temperature \( \Delta T \) is:
\[ \Delta T = T_f - T_i = 150 - 300 = -150 \, K \]
Thus:
\[ Q_1 = 2 \, \text{moles} \cdot C_v \cdot (-150) \]

### Step 4: Heating at Constant Pressure
Now, the gas is heated at constant pressure until it reaches the initial temperature \( T_i = 300 \, K \). The temperature change during this process is:
\[ \Delta T = T_i - T_f = 300 - 150 = 150 \, K \]
The heat absorbed during this process \( Q_2 \) can be calculated using:
\[ Q_2 = n C_p (T_i - T_f) \]
Substituting the values:
\[ Q_2 = 2 \, \text{moles} \cdot C_p \cdot 150 \]

### Step 5: Total Heat Absorbed
The total heat absorbed \( Q_{total} \) is the sum of the heat absorbed during heating and the heat released during cooling:
\[ Q_{total} = Q_1 + Q_2 \]
Substituting the expressions for \( Q_1 \) and \( Q_2 \):
\[ Q_{total} = 2 C_v (-150) + 2 C_p (150) \]
Using the relation \( C_p = C_v + R \):
\[ Q_{total} = 2 C_v (-150) + 2 (C_v + R)(150) \]
This simplifies to:
\[ Q_{total} = 2 C_v (-150) + 2 C_v (150) + 2R(150) \]
The \( C_v \) terms cancel out:
\[ Q_{total} = 2R(150) = 300R \]

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