Question

One mole of a diatomic gas is expanding isothermally from V to 2V at 27°C. If the magnitude of work done by gas in this case is same as the work done in an adiabatic process where initial temperature is 27°C and final temperature is T. Find T in °C.

• A. -37°C
• B. -57°C
• C. -35°C
• D. -55°C

Hint:

Always remember to use absolute temperature (Kelvin) in all thermodynamic calculations.
For adiabatic processes, the work done formula \(W = (P_iV_i - P_fV_f)/(\gamma-1)\) is equivalent to \(W = nR(T_i - T_f)/(\gamma-1)\). Use the one that is most convenient for the given information.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to compare the work done in two different thermodynamic processes: an isothermal expansion and an adiabatic expansion. We are given that the magnitudes of work are equal, and we need to find the final temperature of the adiabatic process.
Step 2: Key Formula or Approach:
- Work done during an isothermal process: \(W_{iso} = nRT \ln\left(\frac{V_f}{V_i}\right)\).
- Work done during an adiabatic process: \(W_{adia} = \frac{n R (T_i - T_f)}{\gamma - 1}\).
- For a diatomic gas, the ratio of specific heats \(\gamma = \frac{7}{5}\).
Step 3: Detailed Explanation:
Part A: Calculate Work Done in Isothermal Expansion
- Number of moles, \(n = 1\).
- Temperature, \(T_{iso} = 27^\circ\text{C} = 27 + 273 = 300\) K.
- Initial volume \(V_i = V\), Final volume \(V_f = 2V\).
\[ W_{iso} = (1) R (300) \ln\left(\frac{2V}{V}\right) = 300R \ln(2) \] Part B: Calculate Work Done in Adiabatic Expansion
- Number of moles, \(n = 1\).
- Initial temperature, \(T_i = 27^\circ\text{C} = 300\) K.
- Final temperature, \(T_f = T\) (in Kelvin).
- For a diatomic gas, \(\gamma = 7/5\), so \(\gamma - 1 = 2/5\).
\[ W_{adia} = \frac{(1) R (300 - T)}{7/5 - 1} = \frac{R(300 - T)}{2/5} = \frac{5R}{2}(300 - T) \] Part C: Equate the Work Done
Given \(|W_{iso}| = |W_{adia}|\). Since both are expansions, work is positive.
\[ 300R \ln(2) = \frac{5R}{2}(300 - T) \] Cancel R from both sides:
\[ 300 \ln(2) = \frac{5}{2}(300 - T) \] We use the standard approximation \(\ln(2) \approx 0.693\).
\[ 300 \times 0.693 = 2.5 \times (300 - T) \] \[ 207.9 = 750 - 2.5T \] \[ 2.5T = 750 - 207.9 = 542.1 \] \[ T = \frac{542.1}{2.5} \approx 216.84 \text{ K} \] Part D: Convert to Celsius
The final temperature in Celsius is:
\[ T(^\circ\text{C}) = T(\text{K}) - 273 = 216.84 - 273 = -56.16^\circ\text{C} \] Step 4: Final Answer:
The calculated temperature is approximately \(-56.16^\circ\text{C}\), which is closest to the option -57°C.