Question

An ideal gas absorbs 750 J of heat as it performs 625 J of work. What is the resulting change in temperature if there are 1.3 moles of the gas in the system

Answer 1

Answer:

7.7°K

Explanation:

Using pV = NkT = nRT

U = CvNkT = Cv nRT

where:

Cv is a constant dependent on temperature (e.g. equal to 3/2 for a monatomic gas for moderate temperatures)

U is the internal energy

p is the pressure

V is the volume

n is the amount of gas (moles)

R is the gas constant, 8.314 J·K^−1mol^-1

T is the absolute temperature

N is the number of particles

k is the Boltzmann constant, 1.381×10^−23 J·K^−1

In your first problem, the energy difference between heat absorbed by the gas and the work performed is the energy lost to heat. So

∆E = Cv nR∆T

where ∆T is the temperature change

∆T = ∆E / (Cv nR)

= (750-625) / (3/2)(1.3)(8.314)

= 7.71° K