Question

A 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 K. It expands from initial volume 19 L to final volume V2. A total of 1.7 kJ of heat is added during the expansion process. What is V2? Let the ideal-gas constant R = 8.314 J/(mol • K).

32 L

41 L

26 L

35 L

Answer 1

Answer:

26 L

Explanation:

According to the first law of thermodynamics, for an ideal gas:

$\Delta U=Q-W$

where

$\Delta U$ is the change in internal energy of the gas

Q is the heat absorbed by the gas

W is the work done by the gas

The internal energy of a gas depends only on its temperature. Here the temperature of the gas is kept constant (330 K), so the internal energy does not change, therefore

$\Delta U=0$

So we have

$Q=W$

The heat added to the gas is

$Q=1.7 kJ = 1700 J$

So this is also equal to the work done by the gas:

$W=1700 J$

For a process at constant temperature, the work done by the gas is given by

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

where:

n is the number of moles

R is the gas constant

T is the temperature of the gas

$V_1$ is the initial volume

$V_2$ is the final volume

In this problem, we have:

W = 1700 J is the work done by the gas

n = 2.00 mol

T = 300 K is the gas temperature

$V_1=19 L$ is the initial volume of the gas

And solving the equation for V2, we find the final volume of the gas:

$V_2=V_1 e^{\frac{W}{nRT}}=(19)e^{\frac{1700}{(2.0)(8.314)(330)}}=26 L$