Two Questions on Specific Heat of Gases

Questions involving specific heats of gases have been discussed in some of the posts on this site. By clicking on the label ‘molar specific heat’ below this post, you will find some of the useful posts in this context. In particular, go through the post dated 1^st September 2006.

Let us discuss two more questions involving specific heats of gases:

(1) An ideal diatomic gas in a container is heated so that half of the gas molecules dissociate into atoms. The molar specific heats (at constant volume) of the sample of the gas in the container before and after heating are C₁ and C₂. Then C₁/C₂ is

(a) 3/7 (b) 5/7 (c) 7/9

(d) 9/10 (e) 15/11

The important point to note here is that on dissociation, each particle (diatomic molecule) with 5 degrees of freedom produces two particles (individual atoms) with 3 degrees of freedom. Therefore you have to use the value C_v = (5/2)R for the undissociated molecule and the value Cv = (3/2)R for the atoms formed on dissociation (C_v is the molar specific heat at constant volume and R is the universal gas constant).

Assuming that there are ‘n’ moles of the diatomic gas initially, the number of moles after dissociation is (3/2)n, with (n/2)×2 = n moles of atoms and n/2 moles of molecules.

The molar specific heat (at constant volume) before dissociation,

C₁ = (5/2)R, appropriate for a diatomic gas.

The molar specific heat (at constant volume) after dissociation,

C₂ = (Heat supplied for increasing the temperature through 1 K) /Number of moles

= [n×(3/2)R + (n/2)×(5/2)R] / (3/2)n = (11/6)R

Therefore, C₁/C₂ = (5/2)R/(11/6)R = 15/11.

(2) An ideal diatomic gas is heated at constant pressure. What is the fraction of the heat energy supplied, which increases the internal energy of the gas?

(a) 2/3 (b) 3/5 (c) 5/7

(d) 7/9 (e) 1/2

This is a simple question but it has appeared in various entrance test papers.

When you heat a gas at constant pressure, the gas expands, thereby doing work. When one mole of a diatomic gas is considered, the increase in the internal energy on heating it through 1 K is equal to its molar specific heat (molar heat capacity) at constant volume, which is (5/2) R where R is the universal gas constant. The total heat energy supplied in this case for increasing the internal energy and for doing the external work is the molar specific heat at constant pressure, which is (7/2) R. The fraction required in the question is therefore [(5/2)R] /[(7/2)R] = 5/7

Change in Internal Energy of a Heated sphere - Two Questions

If you heat a body, its temperature rises and so its internal energy increases. You have come across the increase in internal energy of a gas on many occasions and have noted the difference between the amounts of heat to be supplied on heating the gas at constant volume and at constant pressure to undergo a given temperature rise. Here are two questions involving the work done by a solid on heating it:

(1) If a solid of volume V is heated at constant pressure so as to have a small temperature rise, the work done by the solid will be directly proportional to

(a) V^1/2 (b) V^–1/2 (c) V (d) V^–1 (e) V⁰

(2) The temperature of a copper sphere of volume ‘V’m³ and density ‘ρ’ kg m^–3 increases by a small value ∆T when it absorbs a small quantity ∆Q joule of heat at atmospheric pressure ‘P’ pascal. If the specific heat of copper is ‘C’ J kg^–1K^–1, and the cubical expansivity of copper is ‘γ’ K^–1, the increase in internal energy of the copper sphere is

(a) ∆Q–PV

(b) ∆Q

(c) ∆Q(1– γPV/ρC)

(d) ∆Q(1– γP/ρC)

(e) ∆Q(1– γP/VρC)

Let us consider the second question first since it will give us the answer for the first question also.

The entire heat energy absorbed by the sphere is not used in increasing its internal energy. A small portion is used to do work in expanding (against the atmospheric pressure). Normally we omit this small amount of energy since the expansion of a solid is small. But you have to calculate it here since you are given all data for calculating it.

The rise in temperature of the sphere ∆T = ∆Q/mC where ‘m’ is the mass of the sphere.

Increase in volume of the sphere ∆V = Vγ∆T = (m/ρ)γ(∆Q/mC) = γ∆Q/ρC. [The density ρ can be assumed to be constant as the temperature rise is small].

Therefore, work done against the atmospheric pressure = P∆V = Pγ∆Q/ρC.

The increase in internal energy of the sphere = ∆Q – (Pγ∆Q/ρC) = ∆Q(1– γP/ρC).

So, the correct option is (d).

Now the answer for question No.1 is option (e) since the above expression for increase in internal energy does not contain the volume V.