A piston/cylinder assembly contains butane, C4 H10, at 300^∘ C and 100 kPa with a volume of 0.02

step 1 From here we have in part A, T of RL is equal to T upon TC, which is equals to 573 .15 divided b

A piston/cylinder assembly contains butane, C4 H10, at 300^∘...

A piston/cylinder assembly contains butane, $\mathrm{C}_{4} \mathrm{H}_{10},$ at $300^{\circ} \mathrm{C}$ and $100 \mathrm{kPa}$ with a volume of $0.02 \mathrm{m}^{3} .$ The gas is now compressed slowly in an isothermal process to $300 \mathrm{kPa}$ a. Show that it is reasonable to assume that butane behaves as an ideal gas during this process. b. Determine the work done by the butane during the process.

A piston/cylinder assembly contains butane, $\mathrm{C}_{4} \mathrm{H}_{10},$ at $300^{\circ} \mathrm{C}$ and $100 \mathrm{kPa}$ with a volume of
$0.02 \mathrm{m}^{3} .$ The gas is now compressed slowly in an isothermal process to $300 \mathrm{kPa}$
a. Show that it is reasonable to assume that butane behaves as an ideal gas during this process.
b. Determine the work done by the butane during the process.

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Key Concepts

Ideal Gas Assumption

Ideal gases obey the ideal gas law (PV = nRT) with negligible intermolecular forces and molecular volumes. The assumption of ideal gas behavior is considered reasonable at sufficiently high temperatures and low pressures, where interactions among molecules are minimal. In practical applications, when a gas is at elevated temperatures and not near its condensation conditions, deviations from ideal behavior are minimal, thus justifying the use of the ideal gas model.

Isothermal Process

An isothermal process is one in which the temperature remains constant throughout. For an ideal gas undergoing an isothermal process, the internal energy, which depends solely on temperature, remains constant. This implies that any work done by or on the system must be balanced by a heat transfer to keep the temperature unchanged.

Work in an Isothermal Process

The work done during an isothermal process for an ideal gas can be calculated by integrating the pressure with respect to volume. This leads to the equation W = nRT ln(V?/V?) or equivalently, using pressures, W = nRT ln(P?/P?) since nRT remains constant. This formulation arises from the reversible nature of the process and the logarithmic relationship stems from integrating 1/V with respect to V.

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