Two moles of an ideal monoatomic gas are confined within a cylinder by a massless and frictionless spring loaded piston of cross-sectional area `4 xx 10^(-3)m^(2)`. The spring is, initially in its relaxed state. Now the gas is heated by an electric heater, placed inside the cylinder, for some time. During this time, the gas expands and does `50J` of work in moving the piston through a distance `0.10m`. The temperature of the gas increases by `50K`. Calculate the spring constant and the heat supplied by the heater. `P_(atm) = 1 xx 10^(5)N//m^(2)R = 8.314 J//mol-K`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the heat supplied by the heater (Q)

The heat absorbed by the system can be calculated using the formula:
\[ Q = n C_p \Delta T \]

Where:
- \( n = 2 \) moles (given)
- \( C_p \) for a monoatomic gas is \( \frac{5}{2} R \)
- \( R = 8.314 \, J/(mol \cdot K) \)
- \( \Delta T = 50 \, K \)

Substituting the values:
\[ C_p = \frac{5}{2} \times 8.314 = 20.785 \, J/(mol \cdot K) \]
\[ Q = 2 \times 20.785 \times 50 = 2078.5 \, J \]

### Step 2: Calculate the change in internal energy (U)

The change in internal energy can be calculated using the formula:
\[ U = n C_v \Delta T \]

Where:
- \( C_v \) for a monoatomic gas is \( \frac{3}{2} R \)

Substituting the values:
\[ C_v = \frac{3}{2} \times 8.314 = 12.471 \, J/(mol \cdot K) \]
\[ U = 2 \times 12.471 \times 50 = 1247.1 \, J \]

### Step 3: Apply the first law of thermodynamics

According to the first law of thermodynamics:
\[ Q = U + W \]

We can rearrange this to find W:
\[ W = Q - U \]
Substituting the values:
\[ W = 2078.5 - 1247.1 = 831.4 \, J \]

### Step 4: Determine the work done on the spring

The total work done (W) is the sum of the work done by the gas and the work done on the spring:
\[ W = W_{\text{gas}} + W_{\text{spring}} \]
Where:
- \( W_{\text{gas}} = 50 \, J \) (given)

Thus:
\[ W_{\text{spring}} = W - W_{\text{gas}} = 831.4 - 50 = 781.4 \, J \]

### Step 5: Relate the work done on the spring to the spring constant (k)

The work done on the spring can be expressed as:
\[ W_{\text{spring}} = \frac{1}{2} k x^2 \]

Where:
- \( x = 0.1 \, m \) (displacement)

Substituting the values:
\[ 781.4 = \frac{1}{2} k (0.1)^2 \]
\[ 781.4 = \frac{1}{2} k (0.01) \]
\[ k = \frac{781.4 \times 2}{0.01} = 15628 \, N/m \]

### Final Answers:
- The spring constant \( k \) is \( 15628 \, N/m \).
- The heat supplied by the heater \( Q \) is \( 2078.5 \, J \).

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