Question

Suppose a tank contains 687 m³ of neon (Ne) at an absolute pressure of 1.01×10⁵ Pa. The temperature is changed from 293.2 to 295.9 K. What is the increase in the internal energy of the neon?

Answer 1

We know that increase in internal energy is given by:

dU = U2 - U1 = n*Cv*dT

dU = n*Cv*(T2 - T1)

Cv for monoatomic gas = 3R/2

dU = 3*n*R*(T2 - T1)/2

Now we know from ideal gas law that

PV = nRT

n*R = P*V/T

So,

dU = 3*P1*V1*(T2 - T1)/(2*T1)

Using given values:

dU = 3*1.01*10^5*687*(295.9 - 293.2)/(2*293.2)

dU = 958449.352 J = 958.449 kJ