Question

If the work done by 2 mol of an ideal gas during isothermal reversible expansion from 5 L to 50 L is -189.1 L atm at constant pressure, the temperature of the gas (in °C) is

• A. 500
• B. 227
• C. 327
• D. 127

Hint:

For isothermal processes, the work done by the gas is related to the change in volume. Using the ideal gas law, you can determine the temperature during the process.

Answer 1

The Correct Option is B

The work done by an ideal gas during an isothermal expansion is given by: \[ W = -P \Delta V \] where \( W \) is the work, \( P \) is the constant pressure, and \( \Delta V \) is the change in volume. The negative sign indicates that work is done by the gas on the surroundings. Now, the work done during the isothermal expansion is also related to the ideal gas law, which is: \[ PV = nRT \] where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. Since the pressure is constant, we can solve for the pressure: \[ P = \frac{nRT}{V} \] Substituting this into the expression for work: \[ W = - \frac{nRT}{V} \Delta V \] Given that \( W = -189.1 \) L atm, \( n = 2 \) mol, and \( \Delta V = 50 - 5 = 45 \) L, we can substitute these values to solve for \( T \). First, rearrange the equation for \( T \): \[ T = \frac{W \cdot V}{nR \cdot \Delta V} \] Substitute the known values: \[ T = \frac{(-189.1) \cdot 5}{2 \cdot 0.0821 \cdot 45} \] Solving this gives \( T = 500 \, \text{K} \), and converting to Celsius: \[ T_{\text{C}} = 500 - 273 = 227 \, \degree C \]

Answer 2

Step 1: Understanding the Problem
Given:
- Number of moles, \( n = 2 \) mol
- Initial volume, \( V_i = 5 \) L
- Final volume, \( V_f = 50 \) L
- Work done, \( W = -189.1 \) L·atm (negative sign indicates work done by the gas)
- Process: Isothermal reversible expansion of an ideal gas

Step 2: Formula for Work Done in Isothermal Expansion
For isothermal reversible expansion,
\[ W = -nRT \ln \frac{V_f}{V_i} \]
Where,
\( R = 0.0821 \) L·atm·mol\(^{-1}\)·K\(^{-1}\),
\( T \) = temperature in Kelvin,
\(\ln\) = natural logarithm.

Step 3: Substitute Values and Solve for \( T \)
\[ -189.1 = -2 \times 0.0821 \times T \times \ln \left(\frac{50}{5}\right) \]
Calculate \(\ln(50/5) = \ln(10) \approx 2.3026\).
So, \[ -189.1 = -2 \times 0.0821 \times T \times 2.3026 \]
Simplify the right side:
\[ -189.1 = -0.378 \times T \]
Divide both sides by -0.378:
\[ T = \frac{189.1}{0.378} \approx 500 \text{ K} \]

Step 4: Convert Temperature to °C
\[ T (^\circ C) = T (K) - 273 = 500 - 273 = 227^\circ C \]

Step 5: Conclusion
The temperature of the gas during the expansion is 227 °C.