Inverse: from L to volume ratio

Problem

One mole of gas at $T = 400\ \text{K}$ does $L = 1500\ \text{J}$ of work during an isothermal expansion. Calculate the ratio $V_2/V_1$.

Solution

Isothermal work formula. For the isothermal process

$L = nRT\ln\frac{V_2}{V_1}$

Inversion. Let us isolate the logarithm:

$\ln\frac{V_2}{V_1} = \frac{L}{nRT} = \frac{1500}{8{,}31\cdot 400} = \frac{1500}{3324} = 0{,}4513$

Exponential. Taking the exponential:

$\frac{V_2}{V_1} = e^{0{,}4513} = 1{,}570$

Conclusion. The gas has expanded by about $57\%$:

$\ev{V_2/V_1\approx 1{,}57}$

Links

Topics: Thermodynamics Concepts: Thermodynamic processes Skills: Symbolic setup Objects: Ideal gas