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

26 L

Explanation:

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

where

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

So we have

The heat added to the gas is

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

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

where:

n is the number of moles

R is the gas constant

T is the temperature of the gas

is the initial volume

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

is the initial volume of the gas

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

an obstruction

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y