Question

A monatomic ideal gas expands slowly to twice its original volume, doing 450 J of work in the process. Find the heat added to the gas and the change in internal energy of the gas if the process is (a) isothermal; (b) adiabatic; (c) isobaric.

Short answer

(a)the change in internal energy of the gas if the process is isothermal isQ=450JandΔU=0

(b)the change in internal energy of the gas if the process is adiabatic isQ =0andΔU= -450 J

(c)the change in internal energy of the gas if the process is isobaric is Q = 1125 J and ΔU = 675 J

Step-by-step solution

Step 1:

ideal gas expands to twice its original volume i.e., V2=2V1

work done W= 450 J.

calculate change in internal energy for isothermal, adiabatic, isobaric processes by using first law of thermodynamics.

Use,Q=W+ΔU

Step 2:

(a) the change in internal energy of the gas if the process is isothermal

or the the temperature is constant during the isothermal process .so,ΔT= 0. change in internal energy depends directly onΔT, ΔUequals zero by

$∆U=n{C}_V∆T$ (1)

put the values for W and ΔU into equation (1)

$$\begin{gathered} Q=W+∆U \\ Q=450+0 \\ Q=450J \end{gathered}$$

In isothermal process all heat goes to increase the work done.

Thus, the change in internal energy of the gas if the process is isothermal is Q=450J and ΔU=0

Step 3:

(b)the change in internal energy of the gas if the process is adiabatic.

here the gas is insulated and no heat added to the system so heat Q,

Q=0

put our values for W and into equation (1)

$$\begin{gathered} ∆U=Q-W \\ ∆U=0-450 \\ ∆U=-450J \end{gathered}$$

here the added heat is zero, and the energy to do work comes from the internal energy that decreases.

Thus, the change in internal energy of the gas if the process is adiabatic is Q =0 and ΔU = -450 J

 

(c) the change in internal energy of the gas if the process is isobaric.

in isobaric process the pressure is constant and heat is given by,

$Q=n{C}_P∆T$ (2)

n is the number of moles and Cp is the molar heat capacity at constant pressure and for the monatomic gas Cp=25R where RRis the gas constant.

calculate the heat Q by excluding the term nΔT their values are not there. so the work done is,

$$\begin{gathered} W=P∆V \\ W=nR∆T \\ ∆T=\frac{W}{nR} \end{gathered}$$

put the value of ΔT in equation (2)

$$\begin{gathered} Q=n{C}_P∆T \\ Q=n\left(\frac{5}{2}R\right)\frac{W}{nR} \\ Q=\frac{5}{2}\times 450J \\ Q=1125J \end{gathered}$$

use these values for Q and W and put it into equation (1),

$$\begin{gathered} ∆U=Q-W \\ ∆U=1125J-450J \\ ∆U=675J \end{gathered}$$

The heat added in the isobaric process goes to increase the work done by the gas and its internal energy.

Thus, the change in internal energy of the gas if the process is isobaric is Q = 1125 J and ΔU = 675 JQ