Question

One mole of an ideal gas at 300 K and 20 atm expands to 2 atm under isothermal and reversible conditions. The work done by the gas is \(-x\ \text{kJ mol}^{-1}\). The value of \(x\) is
\((R=8.3\ \text{J K}^{-1}\ \text{mol}^{-1})\)

• A. 5.73
• B. 7.37
• C. 3.75
• D. 4.57

Hint:

Remember the value \(2.303 RT\) at 300 K is approximately \(5700 \text{ J}\) or \(5.7 \text{ kJ}\). Since \(\log(P_1/P_2) = \log(10) = 1\), the answer is directly related to this factor.

Answer 1

The Correct Option is A

Step 1: Formula for Reversible Isothermal Expansion Work \[ w = -2.303 nRT \log_{10} \left( \frac{P_1}{P_2} \right) \] Given: \(n = 1\ \text{mol}\) \(T = 300\ \text{K}\) \(R = 8.3\ \text{J K}^{-1} \text{mol}^{-1}\) \(P_1 = 20\ \text{atm}\) \(P_2 = 2\ \text{atm}\)
Step 2: Calculate work done \[ w = -2.303 \times 1 \times 8.3 \times 300 \times \log_{10} \left( \frac{20}{2} \right) \] \[ w = -2.303 \times 2490 \times \log_{10}(10) \] Since \(\log_{10}(10) = 1\): \[ w = -2.303 \times 2490 \] \[ w \approx -5734.47\ \text{J} \]
Step 3: Convert to kJ and find \(x\) \[ w \approx -5.73\ \text{kJ} \] Given \(w = -x\ \text{kJ}\). \[ x = 5.73 \] Final Answer:
5.73.