Question

The temperature of \(2\) moles of a monoatomic gas changes from \(25^{\circ}C\) to \(35^{\circ}C\) in an adiabatic process. Find the work done.

• A. \(-249.4\,J\)
• B. \(+249.4\,J\)
• C. \(-498.8\,J\)
• D. \(+498.8\,J\)

Hint:

For an adiabatic process, \[ \boxed{Q=0} \] and \[ \boxed{W=-\Delta U.} \] For a monoatomic ideal gas, \[ \boxed{C_V=\frac{3R}{2}} \] Hence, \[ \boxed{W=-\frac32nR\Delta T.} \] Always calculate the temperature difference in Kelvin. Since temperature intervals are the same in Celsius and Kelvin, \[ \boxed{\Delta T(^\circ C)=\Delta T(K).} \]

Answer 1

The Correct Option is A

Concept: An adiabatic process is a thermodynamic process in which there is no heat exchange between the system and the surroundings. Therefore, \[ \boxed{Q=0} \] According to the First Law of Thermodynamics, \[ Q=\Delta U+W, \] where \[ Q=\text{Heat supplied}, \] \[ \Delta U=\text{Change in internal energy}, \] \[ W=\text{Work done by the gas}. \] Since \(Q=0\), \[ \boxed{W=-\Delta U.} \] For a monoatomic ideal gas, \[ \boxed{\Delta U=nC_V\Delta T} \] where \[ C_V=\frac{3R}{2}. \] Hence, \[ \boxed{W=-nC_V\Delta T =-\frac{3}{2}nR\Delta T.} \]

Step 1: Write the given data.
Given, \[ n=2\;\text{moles}, \] \[ T_1=25^{\circ}C, \] \[ T_2=35^{\circ}C. \] Therefore, \[ \Delta T = T_2-T_1 = 35-25 = 10\,K. \] Also, \[ R=8.314\,J\,mol^{-1}K^{-1}. \]

Step 2: Calculate the change in internal energy.
For a monoatomic gas, \[ \Delta U = \frac32 nR\Delta T. \] Substituting the given values, \[ \Delta U = \frac32\times2\times8.314\times10. \] Since, \[ \frac32\times2=3, \] we get \[ \Delta U = 3\times8.314\times10. \] \[ \Delta U = 249.42\,J. \] Thus, \[ \boxed{\Delta U=249.4\,J.} \]

Step 3: Determine the work done.
Since the process is adiabatic, \[ W=-\Delta U. \] Hence, \[ W = -249.4\,J. \] Therefore, \[ \boxed{W=-249.4\,J.} \] Thus, the correct answer is \[ \boxed{\textbf{Option (A)}}. \]

Important Observation: Using the standard thermodynamic convention, \[ \boxed{W=-nC_V\Delta T,} \] the work done by the gas is \[ \boxed{-249.4\,J.} \] Therefore, Option (A) is mathematically correct.