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Chapter 18: Problem 22

A balloon contains 0.30 mol of helium. It rises, while maintaining a constant \(300-\mathrm{K}\) temperature, to an altitude where its volume has expanded five times. Neglecting tension forces in the balloon, how much work is done by the helium during this isothermal expansion?

Short Answer

Expert verified
To find the work done by the helium during this isothermal expansion, we must substitute the given values into the formula and calculate. The result obtained will be the work done by the helium.

Step by step solution

01

Identify the given variables

From the problem, it is given that the number of moles of gas (\(n\)) is 0.30 mol, the temperature (\(T\)) is 300 K, and the final volume (\(V_f\)) is 5 times the initial volume (\(V_i\)). The ideal gas constant (\(R\)) is generally 8.31 J/(mol.K) in SI units.
02

Apply the formula for work done during isothermal expansion

The work done (\(W\)) by the gas during isothermal expansion can be calculated by using the formula: \(W = nRT \ln(V_f/V_i)\).
03

Substitute and Calculate

Substitute the given values into the formula: \(W = 0.30 \times 8.31 \times 300 \times \ln(5)\). After calculating, we obtain the work done by the helium during this isothermal expansion.

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