Question

What is the constant external pressure of an ideal gas when expanded from $2 \times 10^{-2}\ \text{m}^3$ to $3 \times 10^{-2}\ \text{m}^3$, if the work done by the gas is $-5.09\ \text{kJ}$?

• A. $5.09 \times 10^5\ \text{Nm}^{-2}$
• B. $1.01 \times 10^5\ \text{Nm}^{-2}$
• C. $2.02 \times 10^5\ \text{Nm}^{-2}$
• D. $5.60 \times 10^5\ \text{Nm}^{-2}$

Hint:

Always ensure your units are perfectly homogeneous before multiplying or dividing. Converting kJ straight to Joules (J) ensures that your pressure output yields directly in standard pascals or $\text{Nm}^{-2}$ without messy conversion errors at the end!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem describes an expansion of an ideal gas against a constant external pressure. We are given the initial volume, final volume, and the work performed by the system, and we need to solve for the value of that constant external pressure.

Step 2: Key Formula or Approach:
The work done during irreversible expansion against a constant external pressure ($P_{\text{ext}}$) is expressed by the formula:
$$W = -P_{\text{ext}}\Delta V = -P_{\text{ext}}(V_2 - V_1)$$

Step 3: Detailed Explanation:
Let's list the given parameters with proper SI units:
Initial Volume ($V_1$) = $2 \times 10^{-2}\ \text{m}^3$
Final Volume ($V_2$) = $3 \times 10^{-2}\ \text{m}^3$
Work Done ($W$) = $-5.09\ \text{kJ} = -5090\ \text{J}$
First, calculate the volume change ($\Delta V$):
$$\Delta V = V_2 - V_1 = (3 \times 10^{-2}) - (2 \times 10^{-2}) = 1 \times 10^{-2}\ \text{m}^3$$ Now, substitute the values of $W$ and $\Delta V$ into the work formula to solve for $P_{\text{ext}}$:
$$-5090 = -P_{\text{ext}} \times (1 \times 10^{-2})$$ Cancel out the negative signs from both sides of the equation:
$$5090 = P_{\text{ext}} \times 10^{-2}$$ Isolate $P_{\text{ext}}$ by dividing both sides by $10^{-2}$ (which is equivalent to multiplying by $10^2$):
$$P_{\text{ext}} = \frac{5090}{10^{-2}} = 5090 \times 10^2 = 509000\ \text{Nm}^{-2}$$ Expressing this value in standard scientific notation gives:
$$P_{\text{ext}} = 5.09 \times 10^5\ \text{Nm}^{-2}$$ This strictly matches the value provided in option (A).

Step 4: Final Answer:
The constant external pressure is $5.09 \times 10^5\ \text{Nm}^{-2}$, which corresponds to option (A).