Temperature of 5 moles of a gas a decreased by 2K at constant pressure. Indicate the correct statement.

Options

A. work done by the gas =5R
B. work done by the gas =10R
C. work done over the gas =10R
D. work done =0

Explanation

To solve the problem, we need to analyze the situation using the ideal gas law and the concept of work done by or on the gas during a temperature change at constant pressure. Here’s a step-by-step breakdown:

### Step 1: Understand the Given Information
We have:
- Number of moles of gas (n) = 5 moles
- Change in temperature (ΔT) = -2 K (decrease in temperature)
- Pressure (P) = constant

### Step 2: Use the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

### Step 3: Initial and Final Temperatures
Let’s denote the initial temperature as \( T_1 \) and the final temperature as \( T_2 \):
- \( T_1 = T \)
- \( T_2 = T - 2 \)

### Step 4: Calculate Initial and Final States
Using the ideal gas law:
- Initial state:
\[ PV_1 = nRT_1 \]
\[ PV_1 = 5RT \]

- Final state:
\[ PV_2 = nRT_2 \]
\[ PV_2 = 5R(T - 2) = 5RT - 10R \]

### Step 5: Calculate Work Done
The work done (W) by the gas during an isobaric (constant pressure) process can be calculated using the formula:
\[ W = P(V_2 - V_1) \]

Since we know \( PV_1 \) and \( PV_2 \):
\[ W = P(V_2 - V_1) = P\left(\frac{5RT - 10R}{P} - \frac{5RT}{P}\right) \]
\[ W = P\left(\frac{5RT - 10R - 5RT}{P}\right) \]
\[ W = P\left(\frac{-10R}{P}\right) \]
\[ W = -10R \]

### Step 6: Interpret the Sign of Work Done
The negative sign indicates that the work is done by the gas on the surroundings. This means that as the gas expands or does work, it uses energy, which is why the work is negative.

### Conclusion
The correct statement is that the work done by the gas is negative, indicating that work is done by the system (the gas) on the surroundings.