Question

An ideal monatomic gas at a pressure of 2.0×105N/m2 and a temperature of 300 K undergoes a quasi-static isobaric expansion from 2.0×103to4.0×103cm3.

(a) What is the work done by the gas?
(b) What is the temperature of the gas after the expansion?
(c) How many moles of gas are there?
(d) What is the change in internal energy of the gas?
(e) How much heat is added to the gas?

Answer 1

a)
`W= 400 J`

b)
`T_(2)=600 K`

c)
`n=0.16 mol`

d)
`\Delta E_(int)=598 J`

e)
`Q=998 J`

Step-by-step explanation:

a) Let's recall the definition of work.

`W=\int^(Vf)_(V0)PdV`

Because the system is a quasi-static isobaric expansion, P is constant here, therefore:

`W=P\int^(Vf)_(V0)dV=P(V_(f)-V_(0))`

`W=2.0\cdot 10^(5)(4.0\cdot 10^(-3)-2.0\cdot 10^(-3))= 400 [Nm^(2)]`

b) Using the ideal gas equation we have:

`(P_(1)V_(1))/(T_(1))=nR` (1)

and
`(P_(2)V_(2))/(T_(2))=nR` (2)

We can note that n times R is a constant in (1) and (2), so we can equal those equations.

`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` (3)

Let's solve T₂ for (3), let's recall that P₁ = P₂, so they canceled out

`T_(2)=T_(1)\cdot V_(2)/V_(1)`

`T_(2)=300\cdot 4x10^(3)/2x10^(3)=600 K`

c) Using the equation of ideal gas we have:

`(P_(1)V_(1))/(RT_(1))=n`

`n=(P_(1)V_(1))/(RT_(1))=0.16 mol`

d) We can write the internal energy as a function of Cv, and as we know the Cv is 1.5R for a monoatomic gas.

`\Delta E_(int)=n1.5R\Delta T`

`\Delta E_(int)=0.16\cdot 1.5\cdot 8.3 \cdot (600-300)`

`\Delta E_(int)=598 J`

e) Using the first law of thermodynamic, we have:

`\Delta E_(int)=Q-W`

Finally,

`Q=\Delta E_(int)+W=598+400=998 J`

Have a nice day!