Question

A 1.88-mole sample of an ideal gas is contracted at a constant temperature of 298 K. The initial volume is 24.5 L and the amount of work performed on it is 1040 J. What is the final volume? Let the ideal-gas constant R = 8.314 J/(mol • K).
20.2 L

19.6 L

22.5 L

30.6 L

Answer 1

Answer:

19.6 L

Explanation:

In an isothermal process, the temperature of the gas remains constant.

The work done on an ideal gas in an isothermal process is given by the equation:

$W=-nRT ln\frac{V_f}{V_i}$

where

n is the number of moles

R is the gas constant

T is the temperature of the gas

$V_i$ is the initial volume

$V_f$ is the final volume

In this problem, we have:

W = 1040 J is the work done on the gas

n = 1.88 mol

T = 298 K is the gas temperature

$V_i=24.5 L$ is the initial volume

Solving the equation for Vf, we find the final volume:

$V_f=V_i e^{-\frac{W}{nRT}}=(24.5)e^{-\frac{1040}{(1.88)(8.314)(298)}}=19.6 L$