Question

At constant temperature, one mole of an ideal gas of volume 2L expanded to 100 L against an external pressure of 1 atm under reversible conditions. What is the work done (in J)? (1 L atm = 101.3 J; \(\log 5 = 0.7\))

• A. -163.3
• B. -793.2
• C. +793.2
• D. +326.6

Hint:

Expansion work is negative by convention because the system loses energy to the surroundings.

Answer 1

The Correct Option is B

Step 1: Recall reversible isothermal work.
For one mole of an ideal gas expanding reversibly at constant temperature, the work done is \[ W = -nRT \ln\!\left(\frac{V_2}{V_1}\right) \]
Step 2: Replace nRT using the gas law.
Since $PV = nRT$, we can use $nRT = P_1 V_1 = 1 \text{ atm} \times 2 \text{ L} = 2$ L atm at the start.
Step 3: Plug in the volumes.
The gas goes from $2$ L to $100$ L, so $\dfrac{V_2}{V_1} = 50$, giving \[ W = -(2) \times 2.303 \log(50) \text{ L atm} \]
Step 4: Evaluate the logarithm.
With $\log 5 = 0.7$, we get $\log 50 = \log 5 + \log 10 = 0.7 + 1 = 1.7$.
Step 5: Compute the work in L atm.
\[ W = -2 \times 2.303 \times 1.7 = -7.83 \text{ L atm} \]
Step 6: Convert to joules.
Using $1$ L atm $= 101.3$ J, $W = -7.83 \times 101.3 \approx -793.2$ J, which is option 2.
\[ \boxed{-793.2} \]