An ideal gas initially at pressure 1 bar is being compressed from `30m^(3)` to `10m^(3)` volume and its temperature decreases from 320 K to 280 K then find final pressure of gas.

Options

A. 2.625 bar
B. 3.4 bar
C. 1.325 bar
D. 4.5 bar

Explanation

To find the final pressure of the gas after it has been compressed and its temperature has decreased, we can use the ideal gas law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles of gas (constant)
- \( R \) = Ideal gas constant (constant)
- \( T \) = Temperature in Kelvin

Given:
- Initial pressure, \( P_1 = 1 \) bar = \( 100 \) kPa (since 1 bar = 100 kPa)
- Initial volume, \( V_1 = 30 \, m^3 \)
- Initial temperature, \( T_1 = 320 \, K \)
- Final volume, \( V_2 = 10 \, m^3 \)
- Final temperature, \( T_2 = 280 \, K \)

We need to find the final pressure \( P_2 \).

### Step 1: Use the Ideal Gas Law for Initial State
Using the ideal gas law for the initial state:

\[ P_1 V_1 = n R T_1 \]

### Step 2: Use the Ideal Gas Law for Final State
Using the ideal gas law for the final state:

\[ P_2 V_2 = n R T_2 \]

### Step 3: Eliminate \( nR \)
From the two equations, we can eliminate \( nR \):

1. From the initial state:
\[ nR = \frac{P_1 V_1}{T_1} \]

2. Substitute \( nR \) into the final state equation:
\[ P_2 V_2 = \frac{P_1 V_1}{T_1} T_2 \]

### Step 4: Solve for \( P_2 \)
Rearranging the equation to solve for \( P_2 \):

\[ P_2 = \frac{P_1 V_1 T_2}{V_2 T_1} \]

### Step 5: Substitute the Values
Now substituting the known values into the equation:

\[ P_2 = \frac{(100 \, \text{kPa}) (30 \, m^3) (280 \, K)}{(10 \, m^3) (320 \, K)} \]

### Step 6: Calculate \( P_2 \)
Calculating the right-hand side:

\[ P_2 = \frac{100 \times 30 \times 280}{10 \times 320} \]

\[ P_2 = \frac{840000}{3200} \]

\[ P_2 = 262.5 \, \text{kPa} \]

### Final Answer
The final pressure of the gas is:

\[ P_2 = 2.625 \, \text{bar} \]