Question

\(20.0\,\text{dm}^3\) of an ideal gas \(X\) at \(600\) K and \(0.5\) MPa undergoes isothermal reversible expansion until the pressure of the gas becomes \(0.2\) MPa. Which of the following option is correct? (Given: \(\log 2 = 0.3010\), \(\log 5 = 0.6989\))

• A. \(w=-3.9\,\text{kJ},\ \Delta U=0,\ \Delta H=0,\ q=3.9\,\text{kJ}\)
• B. \(w=9.1\,\text{kJ},\ \Delta U=9.1\,\text{kJ},\ \Delta H=0,\ q=0\)
• C. \(w=-9.1\,\text{kJ},\ \Delta U=0,\ \Delta H=0,\ q=9.1\,\text{kJ}\)
• D. \(w=+4.1\,\text{kJ},\ \Delta U=0,\ \Delta H=0,\ q=-4.1\,\text{kJ}\)

Hint:

In isothermal processes of ideal gases, work done is exactly balanced by heat absorbed or released.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For an isothermal process involving an ideal gas, the change in internal energy (\(\Delta U\)) and enthalpy (\(\Delta H\)) are zero because they are functions of temperature only.
Step 2: Key Formula or Approach:
Work done in isothermal reversible expansion:
\[ w = -2.303 nRT \log\left(\frac{P_1}{P_2}\right) \]
Since \(PV = nRT\), we can use \(P_1 V_1\) in place of \(nRT\).
Step 3: Detailed Explanation:
Given: \(V_1 = 20 \text{ dm}^3 = 20 \times 10^{-3} \text{ m}^3\), \(T = 600 \text{ K}\), \(P_1 = 0.5 \text{ MPa} = 0.5 \times 10^6 \text{ Pa}\), \(P_2 = 0.2 \text{ MPa}\).
First, calculate \(nRT\):
\[ nRT = P_1 V_1 = (0.5 \times 10^6) \times (20 \times 10^{-3}) = 10,000 \text{ J} = 10 \text{ kJ} \]
Now, calculate work (\(w\)):
\[ w = -2.303 \times (10 \text{ kJ}) \times \log\left(\frac{0.5}{0.2}\right) \]
\[ w = -23.03 \times \log(2.5) \]
\[ \log(2.5) = \log(5/2) = \log 5 - \log 2 = 0.6989 - 0.3010 = 0.3979 \]
\[ w = -23.03 \times 0.3979 \approx -9.16 \text{ kJ} \approx -9.1 \text{ kJ} \]
By First Law of Thermodynamics (\(\Delta U = q + w\)):
Since \(\Delta U = 0\), then \(q = -w = +9.1 \text{ kJ}\).
Step 4: Final Answer:
\(w = -9.1 \text{ kJ, } \Delta U = 0, \Delta H = 0, q = 9.1 \text{ kJ}\).