Question

A mole of an ideal gas at pressure 4.00 atm and temperature 298 K expands isothermally to double its volume. What is the work done by the gas?

(a) -2.46 L·atm
(b) -4.92 L·atm
(c) 2.46 L·atm
(d) 4.92 L·atm

Answer 1

The work done by the gas in an isothermal expansion process can be calculated using the formula: work = -nRT ln(Vf/Vi). In this case, the work done is -4.92 L·atm.

Step-by-step explanation:

In an isothermal expansion process, the work done by the gas can be calculated using the formula: work = -nRT ln(Vf/Vi). The work done by the gas in an isothermal expansion process can be calculated using the formula: work = -nRT ln(Vf/Vi). In this case, the work done is -4.92 L·atm.

Where n is the number of moles of the gas, R is the ideal gas constant (0.0821 L·atm/(mol·K)), T is the temperature in Kelvin, Vf is the final volume, and Vi is the initial volume.

In this case, the number of moles of the gas is 1 (as stated in the question) and the initial and final volumes are related by doubling the volume. So, Vi is V and Vf is 2V.

Plugging in the given values, we have: work = -1(0.0821)(298) ln(2) = -4.92 L·atm.