Question

One mole of an ideal gas with heat capacity at constant pressure

C_{p}

, undergoes the process

T = T_{0} + α V

, where

T_{0}

and

α

are constants. Find,

(a)

heat capacity of the gas as a function of its volume,

(b)

the amount of heat transferred to the gas, if its volume is increased from

V_{1}

to

V_{2}

.

Accepted Answer

$(a)$ According to the first law of thermodynamics, the amount of heat given to the system is equal to an increase in internal energy plus the work done by the system.

$d Q = d U + dW d Q = n C_{V} d T + P d V,$

where, $d Q$ is the Heat given to the system,

$d U$ is the increase in internal heat and

$d W$ is the work done by the system.

$P$ is the pressure,

$d V$ is the increase in pressure and

$n$ is the number of moles,

$C_{V}$ is the specific heat of the gas at constant volume.

The molar specific heat is

$C = \frac{d Q}{nd T} . Now, on substituting the value of dQ from above equation and P = \frac{n R T}{V} C = \frac{n C_{V} d T + P d V}{nd T}$

Now on cancelling $n$ , taking common and cancelling $d T$ , we get,
$C = \frac{n C_{V} d T + \frac{n R T}{V} d V}{nd T} = C_{V} + \frac{R T}{V} \frac{d V}{d T}$

As per heat capacity, $C = C_{V} + \frac{R T}{V} \frac{d V}{d T}$ ... $(1)$

Now in the question, the process is given by, $T = T_{0} + α V$

On rearranging the equation, we get,

$V = \frac{T}{α} - \frac{T_{0}}{α}$

After differentiating with respect to $T$ , we get, $\frac{d V}{d T} = \frac{1}{α}$

So now, substituting the value of $\frac{d V}{d T} in equation (1) we get,$

$c = C_{V} + \frac{R T}{V} \frac{1}{α} = \frac{R}{γ - 1} + \frac{R (T_{0} + α V)}{V} \frac{1}{α}$
$= \frac{R}{γ - 1} + R (\frac{T_{0}}{α V} + 1) = \frac{γ R}{γ - 1} + \frac{R T_{0}}{α V} = C_{V} + \frac{R T}{α V} = C_{P} + \frac{R T_{0}}{α V}$

$c = C_{P} + \frac{R T_{0}}{α V}$

$(b)$ Given, the equation of process is $T = T_{0} + α V$

As $T = \frac{p V}{R}$ , for one mole of gas,

$p = \frac{R}{V} (T_{0} + α V) = \frac{R T_{0}}{V} + α R$

Now, the work done by the gas in increasing the volume $V_{1}$ to $V_{2}$ is given by, $W = \int_{V_{1}}^{V_{2}} p d V$ (for one mole $)$
Therefore, Substituting the values in the work done equation, we get, $W = \int_{V_{1}}^{V_{2}} (\frac{R T_{0}}{V} + α R) d V$
$= R T_{0} \ln \frac{V_{2}}{V_{1}} + α (V_{2} - V_{1})$

Now, the internal energy is defined as the energy associated with the random, disordered motion of molecules. It is separated in scale from the macroscopic ordered energy associated with moving objects, it refers to the invisible microscopic energy on the atomic and molecular scale.

And it is given by the equation,

$Δ U = C_{V} (T_{2} - T_{1})$

$= C_{V} [T_{0} + α V_{2} - T_{0} α V_{1}] = α C_{V} (V_{2} - V_{1})$

Now, by the first law of thermodynamics, $Q = Δ U + W$ ,

where, $Q$ is the heat supplied to the system and

$W$ is the work done by the gas.

So now, on substituting the values in the first law of thermodynamics, we get,

$= \frac{α R}{γ - 1} (V_{2} - V_{1}) + R T_{0} \ln \frac{V_{2}}{V_{1}} + α R (V_{2} - V_{1})$

$= α R (V_{2} - V_{1}) [1 + \frac{1}{γ - 1}] + R T_{0} \ln \frac{V_{2}}{V_{1}}$

$= α C_{p} (V_{2} - V_{1}) + R T_{0} \ln \frac{V_{2}}{V_{1}}$
$= α C_{p} (V_{2} - V_{1}) + R T_{0} \ln \frac{V_{2}}{V_{1}}$

I E Irodov Solutions for Chapter: THERMODYNAMICS AND MOLECULAR PHYSICS, Exercise 2: THE FIRST LAW OF THERMODYNAMICS, HEAT CAPACITY

I E Irodov Physics Solutions for Exercise - I E Irodov Solutions for Chapter: THERMODYNAMICS AND MOLECULAR PHYSICS, Exercise 2: THE FIRST LAW OF THERMODYNAMICS, HEAT CAPACITY

Questions from I E Irodov Solutions for Chapter: THERMODYNAMICS AND MOLECULAR PHYSICS, Exercise 2: THE FIRST LAW OF THERMODYNAMICS, HEAT CAPACITY with Hints & Solutions

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