Question

A cylinder contains $5\text{ m}^3$ of an ideal gas at a pressure of 1 bar. The gas is compressed in a reversible isothermal process until the pressure increases to 5 bar. Determine the work required for this compression (kJ).

• A. 804.7
• B. 953.2
• C. 981.7
• D. 1012.2

Hint:

Ensure your pressure unit is in kPa (or $\text{kN/m}^2$) so that multiplying by volume in $\text{m}^3$ yields the work directly in kilojoules (kJ).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This problem requires us to calculate the work done on an ideal gas during a reversible isothermal compression process.

Step 2: Key Formula or Approach:
The work done during a reversible isothermal process for an ideal gas is:
\[ W = P_1 V_1 \ln\left(\frac{V_2}{V_1}\right) \]
Since temperature is constant, Boyle's Law gives $P_1 V_1 = P_2 V_2 \implies \frac{V_2}{V_1} = \frac{P_1}{P_2}$.
Substituting this gives:
\[ W = P_1 V_1 \ln\left(\frac{P_1}{P_2}\right) \]
The magnitude of work required (work input) is:
\[ W_{\text{required}} = P_1 V_1 \ln\left(\frac{P_2}{P_1}\right) \]

Step 3: Detailed Explanation:

• Identify the given parameters:
- Initial volume, $V_1 = 5 \text{ m}^3$
- Initial pressure, $P_1 = 1 \text{ bar} = 100 \text{ kPa}$
- Final pressure, $P_2 = 5 \text{ bar} = 500 \text{ kPa}$

• Substitute the values into the work equation:
\[ W_{\text{required}} = 100 \text{ kPa} \times 5 \text{ m}^3 \times \ln\left(\frac{5}{1}\right) \]
\[ W_{\text{required}} = 500 \times \ln(5) \]

• Using the value of $\ln(5) \approx 1.60944$:
\[ W_{\text{required}} \approx 500 \times 1.60944 = 804.72 \text{ kJ} \]

Step 4: Final Answer:
The work required for this compression is $804.7 \text{ kJ}$.