Question

An ideal gas on isothermal reversible compression from 10 L to 5 L performs 1730 J of work at 300 K. Calculate number of moles of gas involved in compression? ($R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}$)

• A. 2.5
• B. 2
• C. 1
• D. 3

Hint:

The product constant collection $2.303 \times R \times \log_{10}(2) \approx 5.76$. Multiplying $5.76 \times 300\ \text{K}$ yields roughly $1728$. Since the experimental work given is 1730 J, the mole scalar must be exactly 1!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem presents an ideal gas undergoing an isothermal reversible compression from an initial volume ($V_1$) to a final volume ($V_2$) at a specific temperature ($T$). Given the work value ($W$), we are required to determine the total number of moles ($n$) involved in the process.

Step 2: Key Formula or Approach:
The maximum work done during an isothermal reversible compression/expansion of an ideal gas is given by the thermodynamic equation: $$ W = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right) $$ Rearranging the expression to solve for the number of moles ($n$): $$ n = \frac{W}{-2.303 RT \log_{10}\left(\frac{V_2}{V_1}\right)} $$

Step 3: Detailed Explanation:
Let's list the given parameters from the problem statement:

• Initial Volume, $V_1 = 10\ \text{L}$

• Final Volume, $V_2 = 5\ \text{L}$

• Temperature, $T = 300\ \text{K}$

• Gas Constant, $R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}$

• Work performed, $W = 1730\ \text{J}$ (In some textbook conventions for compression work, work done on the system is positive, which cancels out the negative sign from the logarithmic contraction ratio).
Let's evaluate the log ratio first: $$ \log_{10}\left(\frac{5}{10}\right) = \log_{10}(0.5) = -0.3010 $$ Now, substitute these parameters back into the work formula: $$ 1730 = -2.303 \times n \times 8.314 \times 300 \times (-0.3010) $$ Combine the constants in the numerator and denominator: $$ 2.303 \times 8.314 \times 300 = 5744.14 $$ $$ 1730 = n \times 5744.14 \times 0.3010 $$ $$ 1730 \approx n \times 1729 $$ $$ n = \frac{1730}{1729} \approx 1\ \text{mol} $$

Step 4: Final Answer: The number of moles of gas involved in the compression process is 1, which corresponds to option (C).