Question

An ideal gas is brought through an isothermal compression process. The 2.00 moles of gas go from having an initial volume of 219.8 cm3 to 120.5 cm3. If 2182 cal are released by the gas during this process, what are the temperature T of the gas and the final pressure pf?

Answer 1

First Law:

ΔU = ΔQ + ΔW

where ΔU is the increase in internal energy, ΔQ the increase in heat ( i.e. the heat added to the system) and ΔW the work done on the system.

For an ideal gas, U is a function only of temperature. Therefore for an isothermal process ΔU=0. Hence ΔW = -ΔQ.

The work done ON the system in an expansion is

ΔW = - ∫ p dV = -nRT ln(Vf/Vi)

where of course we used p V = n R T to evaluate the integral.

Therefore we obtained

ΔQ = nRT ln(Vf/Vi)

T = ΔQ / (n R ln(Vf/Vi) )

Here this calculates to

T = - 2182J / ( 2.00 mol * 8.31 J/(mol K) * ln(120.5/219.8) )

= 218.42 K


Knowing T, pf follows from

pf= nRT/Vf.

= 2.00mol * 8.31J/(mol K) * 218.4K / (120.5*10^-6 m^3)

= 3.0122*10^7 N/m^2

= 3.0122*10^7 Pa