To solve the given problem step by step, we will break it down into two parts as per the question requirements:
### Part (a): Work Done by the Gas in the Process A to B
1. **Identify the Process**: The process from A to B is an adiabatic expansion. For an adiabatic process, the work done \( W \) can be calculated using the formula:
\[
W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1}
\]
where \( P_1 \) and \( P_2 \) are the pressures at states A and B, \( V_1 \) and \( V_2 \) are the volumes at states A and B, and \( \gamma \) is the heat capacity ratio.
2. **Use the Ideal Gas Law**: Since we do not have the volumes directly, we can use the ideal gas law \( PV = nRT \) to express the volumes in terms of pressure and temperature:
\[
V = \frac{nRT}{P}
\]
Thus,
\[
V_A = \frac{nRT_A}{P_A}, \quad V_B = \frac{nRT_B}{P_B}
\]
3. **Calculate the Pressures**: We know that:
- \( T_A = 1000 \, K \)
- \( P_B = \frac{2}{3} P_A \)
4. **Calculate the Temperature at B**: Using the adiabatic relation \( P_1 V_1^\gamma = P_2 V_2^\gamma \):
\[
P_A \left(\frac{nRT_A}{P_A}\right)^\gamma = P_B \left(\frac{nRT_B}{P_B}\right)^\gamma
\]
This simplifies to:
\[
T_A^\gamma P_A^{1-\gamma} = T_B^\gamma P_B^{1-\gamma}
\]
We can solve for \( T_B \) using the ratio of pressures:
\[
T_B = T_A \left(\frac{P_B}{P_A}\right)^{\frac{\gamma - 1}{\gamma}} = 1000 \left(\frac{2}{3}\right)^{\frac{2}{5}} \approx 850 \, K
\]
5. **Calculate Work Done**: Now substituting the values back into the work done formula:
\[
W = \frac{P_B V_B - P_A V_A}{\gamma - 1}
\]
Using \( V_A \) and \( V_B \) from the ideal gas law:
\[
W = \frac{\left(\frac{2}{3} P_A \cdot \frac{nR \cdot 850}{\frac{2}{3} P_A}\right) - \left(P_A \cdot \frac{nR \cdot 1000}{P_A}\right)}{\frac{5}{3} - 1}
\]
Simplifying this gives:
\[
W = \frac{nR(850 - 1000)}{\frac{2}{3}} = \frac{nR(-150)}{\frac{2}{3}} = -225nR
\]
For \( n = 1 \) mole and \( R = 8.31 \, J/(mol \cdot K) \):
\[
W = -225 \cdot 8.31 \approx -1875 \, J
\]
Since work done by the gas is positive during expansion, we take the absolute value:
\[
W \approx 1875 \, J
\]
### Part (b): Heat Lost by the Gas in the Process B to C
1. **Identify the Process**: The process from B to C is a cooling at constant volume (isochoric process). The heat lost \( Q \) can be calculated using:
\[
Q = n C_v \Delta T
\]
where \( C_v \) is the molar heat capacity at constant volume.
2. **Calculate \( C_v \)**: For a monoatomic ideal gas:
\[
C_v = \frac{3}{2} R
\]
3. **Determine the Temperature Change**: We need to find \( T_C \):
Using the relation for constant volume:
\[
\frac{P_B}{T_B} = \frac{P_C}{T_C}
\]
Given \( P_C = \frac{1}{3} P_A \) and substituting \( P_B = \frac{2}{3} P_A \):
\[
T_C = T_B \cdot \frac{P_C}{P_B} = 850 \cdot \frac{1/3}{2/3} = 850 \cdot \frac{1}{2} = 425 \, K
\]
4. **Calculate \( \Delta T \)**:
\[
\Delta T = T_C - T_B = 425 - 850 = -425 \, K
\]
5. **Calculate Heat Lost**:
\[
Q = n C_v \Delta T = 1 \cdot \frac{3}{2} R \cdot (-425)
\]
Substituting \( R = 8.31 \):
\[
Q = \frac{3}{2} \cdot 8.31 \cdot (-425) \approx -5312.25 \, J
\]
The heat lost is approximately \( 5312.25 \, J \).
### Summary of Results:
- (a) Work done by the gas in the process A to B: \( 1875 \, J \)
- (b) Heat lost by the gas in the process B to C: \( 5312.25 \, J \)
