Three moles of a monoatomic gas is subjected to a rise of temperature 400K under isobaric process. If ratio of increase in internal energy and work done by gas is `x/10` then find x...

Options

A. 10
B. 15
C. 25
D. 28

Answer

To solve the problem, we need to find the ratio of the increase in internal energy to the work done by the gas during an isobaric process. Let's break this down step by step.

### Step 1: Identify the given data
We have:
- Number of moles, \( n = 3 \) moles
- Rise in temperature, \( \Delta T = 400 \, K \)
- For a monoatomic gas, the degrees of freedom \( f = 3 \)

### Step 2: Calculate the work done by the gas
In an isobaric process (constant pressure), the work done \( W \) by the gas can be calculated using the formula:
\[ W = P \Delta V \]
From the ideal gas law, we know:
\[ PV = nRT \]
At constant pressure, we can express the work done as:
\[ W = nR \Delta T \]
Substituting the values we have:
\[ W = 3R \times 400 \]

### Step 3: Calculate the change in internal energy
The change in internal energy \( \Delta U \) for a monoatomic gas is given by:
\[ \Delta U = \frac{f}{2} n R \Delta T \]
Substituting the values:
\[ \Delta U = \frac{3}{2} \times 3R \times 400 \]

### Step 4: Simplify the expressions
Now we can simplify both \( W \) and \( \Delta U \):
1. Work done:
\[ W = 3R \times 400 = 1200R \]

2. Change in internal energy:
\[ \Delta U = \frac{3}{2} \times 3R \times 400 = \frac{9}{2} \times 400R = 1800R \]

### Step 5: Calculate the ratio of increase in internal energy to work done
Now we find the ratio:
\[ \frac{\Delta U}{W} = \frac{1800R}{1200R} = \frac{1800}{1200} = \frac{3}{2} \]

### Step 6: Relate the ratio to the given expression
According to the problem, this ratio is given as \( \frac{x}{10} \):
\[ \frac{3}{2} = \frac{x}{10} \]

### Step 7: Solve for \( x \)
Cross-multiplying gives:
\[ 3 \cdot 10 = 2x \implies 30 = 2x \implies x = 15 \]

### Final Answer
Thus, the value of \( x \) is \( \boxed{15} \).

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