A Container having 1 mole of a gas at a temperature `27^(@)` has a movable piston which maintains at constant pressure in container of 1 atm . The gas is compressed until temperature becomes `127^(@)`. The work done is ( C for gas is 7.03 cal / mol – K )

Options

A. 703 j
B. 814 j
C. 121 j
D. 2035 j

Answer

To solve the problem step by step, we will calculate the work done on the gas during the compression process at constant pressure.

### Step 1: Identify Given Values
- Number of moles of gas, \( n = 1 \) mole
- Initial temperature, \( T_1 = 27^\circ C = 300 \, K \) (convert to Kelvin by adding 273)
- Final temperature, \( T_2 = 127^\circ C = 400 \, K \)
- Constant pressure, \( P = 1 \, atm \)
- Gas constant, \( R = 8.314 \, J/(mol \cdot K) \)

### Step 2: Calculate Change in Temperature
The change in temperature, \( \Delta T \), is given by:
\[ \Delta T = T_2 - T_1 = 400 \, K - 300 \, K = 100 \, K \]

### Step 3: Calculate Work Done at Constant Pressure
The work done, \( W \), on the gas at constant pressure can be calculated using the formula:
\[ W = n \cdot R \cdot \Delta T \]
Substituting the known values:
\[ W = 1 \, mol \cdot 8.314 \, J/(mol \cdot K) \cdot 100 \, K \]

### Step 4: Perform the Calculation
Calculating the work done:
\[ W = 8.314 \cdot 100 = 831.4 \, J \]

### Step 5: Convert Joules to Calories
Since the problem states the specific heat capacity in calories, we can convert joules to calories using the conversion factor \( 1 \, cal = 4.184 \, J \):
\[ W = \frac{831.4 \, J}{4.184 \, J/cal} \approx 198.4 \, cal \]

### Final Answer
The work done on the gas is approximately \( 198.4 \, cal \).

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