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 will follow these steps:

### Step 1: Understand the Initial Conditions
We have a cylinder divided into two equal parts, each with a length of 45 cm. The gas in both parts has the same mass, pressure (1 atm), and temperature (300 K).

- Volume of each part (initially) = \( V_1 = V_2 = 45 \times A \) (where A is the cross-sectional area)

### Step 2: Analyze the Heating Process
When the gas in one part is heated, the piston shifts by 5 cm. This means:
- New volume of the heated part (after heating) = \( V_1' = 50 \times A \)
- New volume of the other part (compressed) = \( V_2' = 40 \times A \)

### Step 3: Apply the Ideal Gas Law
The ideal gas law states:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature

### Step 4: Set Up Equations for the Final State
Since the piston is impermeable to heat, the pressure in both chambers will be equal at equilibrium:
\[ P_1' = P_2' \]

Using the ideal gas law for both parts:
1. For the heated part:
\[ P_1' \cdot V_1' = nR \cdot T_1' \]
\[ P_1' \cdot (50A) = nR \cdot T_1' \]

2. For the other part:
\[ P_2' \cdot V_2' = nR \cdot T_2 \]
\[ P_2' \cdot (40A) = nR \cdot 300 \]

### Step 5: Equate Pressures
Since \( P_1' = P_2' \):
\[ P_1' = \frac{nR \cdot T_1'}{50A} \]
\[ P_2' = \frac{nR \cdot 300}{40A} \]

Setting these equal gives:
\[ \frac{nR \cdot T_1'}{50A} = \frac{nR \cdot 300}{40A} \]

### Step 6: Simplify the Equation
Cancel \( nR \) and \( A \) from both sides:
\[ \frac{T_1'}{50} = \frac{300}{40} \]

### Step 7: Solve for \( T_1' \)
Cross-multiplying gives:
\[ 40T_1' = 300 \times 50 \]
\[ T_1' = \frac{15000}{40} = 375 \, \text{K} \]

### Step 8: Calculate the Final Pressure
Using the ideal gas law again for the heated part:
\[ P_1' \cdot 50A = nR \cdot 375 \]
Substituting \( nR \) from the other part:
\[ P_1' \cdot 50A = \frac{nR \cdot 300}{40} \cdot 50 \]
\[ P_1' = \frac{300 \cdot 50}{40 \cdot 50} = \frac{300}{40} = 7.5 \, \text{atm} \]

### Final Results
Thus, the final temperature and pressure of the gas in the heated part are:
- Temperature \( T_1' = 375 \, \text{K} \)
- Pressure \( P_1' = 1.125 \, \text{atm} \)