Question

A monatomic ideal gas expands from 2.00 m³ to 2.95 m³ at a constant pressure of 2.80 ✕ 10⁵ Pa. Find the following.

(a) Find the work done on the gas. J

(b) Find the thermal energy Q transferred into the gas by heat.
J

(c) Find the change in the internal energy of the gas.
J

Answer 1

The ideal gas law is:

P*V = n*R*T

or

P = n*R*T/V

or T = P*V/(n*R)

Initially, we have that:

T_initial = (2.80*10^5 Pa)*(2 m^3)/((1 mol)*(8.314 J/(mol*K))

T_initial = 67,356 K

(a) The work done by the gas is given by:

w = P*ΔV = (2.80*10^5 Pa)*(2.95 - 2) m^3

w = 2.66*10^5 J

This is the work done BY the gas but the question askes work done on the gas.

So, the work done on the gas = -2.66*10^5 J

(b) From gas kinetic theory, one can show that the constant-pressure heat capacity of a monoatomic ideal gas is:

Cp = 5*n*R/2

so for this system,

Cp = 5*(1 mol)*(8.314 J/(mol*K))/2

Cp = 20.785 J/K

Let’s calculate T_final:

P = n*R*T_final/V_final = n*R*T_initial/V_final

Dividing through by n*R, and plugging in the appropriate numbers:

T_final = T_intitial*(V_final/V_initial) = (67,356 K)*(2.95)/(2)

T_final = 99,350 K

To change the temperature by an amount ΔT = T_final - T_initial, one needs to add an amount of thermal energy equal to:

q = ΔT*Cp = (99,350 K – 67,356 K)*(20.785 J/K)

q = 6.65*10^5 J

(c) From the first law, we know that:

ΔE = q - w

So:

ΔE = 6.65*10^5 J - 2.66*10^5 J = 3.99*10^3 J = 3.99*109^5 J