Question

A gas is allowed to expand against a constant external pressure of $2.5\ \mathrm{bar}$ from an initial volume 'x' $\mathrm{L}$ to final volume of $4.5\ \mathrm{L}$. If amount of work done is $5\ \mathrm{dm^3\ bar}$, what is the value of 'x'?

• A. $2.5\ \mathrm{L}$
• B. $4.5\ \mathrm{L}$
• C. $6.5\ \mathrm{L}$
• D. $1.2\ \mathrm{L}$

Hint:

To remember sign conventions easily: expansion always means the system loses energy by doing work on the surroundings, so $W$ is negative. Thus, $\text{Work Magnitude} = P \times \Delta V \implies 5 = 2.5 \times \Delta V \implies \Delta V = 2\ \mathrm{L}$. Since it expanded to $4.5\ \mathrm{L}$, it must have started at $4.5 - 2 = 2.5\ \mathrm{L}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem outlines an expansion process where an ideal gas changes volume against a uniform constant external pressure ($P_{\mathrm{ext}} = 2.5\ \mathrm{bar}$). We are given the final volume ($V_f = 4.5\ \mathrm{L}$) and the work done during expansion ($5\ \mathrm{dm^3\ bar}$), and we need to determine the initial volume $x$.

Step 2: Key Formula or Approach:
The thermodynamic formula for the work done during an expansion against a constant external pressure is: $$W = -P_{\mathrm{ext}}\Delta V = -P_{\mathrm{ext}}(V_f - V_i)$$ By convention in chemistry, work done by the system during expansion is negative: $$W = -5\ \mathrm{dm^3\ bar} = -5\ \mathrm{L\ bar}$$ (since $1\ \mathrm{dm^3} = 1\ \mathrm{L}$).

Step 3: Detailed Explanation:
Let's substitute the given numerical variables directly into our formula: $$-5 = -2.5 \times (4.5 - x)$$ Divide both sides by $-2.5$: $$\frac{-5}{-2.5} = 4.5 - x$$ $$2 = 4.5 - x$$ Rearrange the equation to isolate the variable $x$: $$x = 4.5 - 2$$ $$x = 2.5\ \mathrm{L}$$

Step 4: Final Answer:
The value of the initial volume 'x' is $2.5\ \mathrm{L}$, corresponding to option (A).