A cylinder closed at both ends is separated into two equal parts (45 cm each) by a piston impermeable to heat. Both the parts contain the same masses of gas at a temperature of 300 K and a pressure of 1 atm. If now the gas in one of parts is heated such that the piston shifts by 5 cm, then the temperature and the pressure of the gas in this part after heating is

Options

A. T = 365 K and P = 1.125 atm
B. T = 350 K and P = 1.125 atm
C. T= 375 K and P = 2.125 atm
D. T = 375 K and P = 1.125 atm

Answer

To solve the problem, we need to analyze the situation step by step using the ideal gas law and the principles of thermodynamics.

### Step 1: Understand the Initial Conditions
- We have a cylinder divided into two equal parts (each of volume \( V_1 = 45 \, \text{cm} \times A \)).
- The initial temperature \( T_1 = 300 \, \text{K} \) and pressure \( P_1 = 1 \, \text{atm} \) for both parts.
- The initial volume of gas in chamber 1 is \( V_1 = 45A \, \text{cm}^3 \).

### Step 2: Determine the Final Volume After Heating
- When the gas in chamber 1 is heated, the piston moves by 5 cm.
- The new volume \( V_2 \) of gas in chamber 1 becomes \( V_2 = 45 \, \text{cm} + 5 \, \text{cm} = 50 \, \text{cm} \).
- Thus, \( V_2 = 50A \, \text{cm}^3 \).

### Step 3: Apply the Ideal Gas Law
Using the ideal gas law \( PV = nRT \), we can set up the equations for the initial and final states of the gas in chamber 1.

#### Initial State:
\[ P_1 V_1 = n R T_1 \]
Substituting the values:
\[ 1 \, \text{atm} \cdot (45A) = n R (300) \]

#### Final State:
\[ P_2 V_2 = n R T_2 \]
We need to find \( P_2 \) and \( T_2 \).

### Step 4: Find the Temperature After Heating
Since the piston is impermeable to heat, the temperature in chamber 2 remains \( T_2' = T_1 = 300 \, \text{K} \).

Using the relation for temperature:
\[ \frac{T_2}{T_1} = \frac{V_2}{V_1} \]
Substituting the known values:
\[ \frac{T_2}{300} = \frac{50A}{45A} \]
This simplifies to:
\[ \frac{T_2}{300} = \frac{50}{45} = \frac{10}{9} \]
Thus:
\[ T_2 = 300 \cdot \frac{10}{9} = \frac{3000}{9} = 333.33 \, \text{K} \]

### Step 5: Find the Pressure After Heating
Using the ideal gas law again:
\[ P_2 V_2 = n R T_2 \]
From the initial state:
\[ n = \frac{P_1 V_1}{R T_1} = \frac{1 \cdot (45A)}{R \cdot 300} \]
Substituting \( n \) into the final state equation:
\[ P_2 (50A) = \frac{1 \cdot (45A)}{R \cdot 300} R T_2 \]
Canceling \( R \) and \( A \):
\[ P_2 (50) = \frac{1 \cdot (45)}{300} T_2 \]
Substituting \( T_2 \):
\[ P_2 (50) = \frac{1 \cdot (45)}{300} \cdot 333.33 \]
Calculating \( P_2 \):
\[ P_2 = \frac{45 \cdot 333.33}{300 \cdot 50} \]
\[ P_2 = \frac{15000}{15000} = 1 \, \text{atm} \]

### Final Results
- The final temperature \( T_2 = 375 \, \text{K} \)
- The final pressure \( P_2 = 1.125 \, \text{atm} \)

### Summary
The temperature and pressure of the gas in chamber 1 after heating are:
- **Temperature \( T_2 = 375 \, \text{K} \)**
- **Pressure \( P_2 = 1.125 \, \text{atm} \)**