Question

Nitrogen gas in an expandable container is cooled from $\mathbf{50}\mathbf{°}\mathit{C}$ to $\mathbf{10}\mathbf{°}\mathit{C}$with the pressure held constant at $\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{P}\mathit{a}$. The total heat liberated by the gas is $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{J}$. Assume that the gas may be treated as ideal. Find (a) the number of moles of gas; (b) the change in internal energy of the gas; (c) the work done by the gas. (d) How much heat would be liberated by the gas for the same temperature change if the volume were constant?

Short answer

a) The number of moles, $n=21.5moles$

b) The change in internal energy, $∆U=-17.9KJ$

c) The amount of work done by the gas, $W=-7.1kJ$

d) The amount of heat, $Q=-17.9kJ$

Step-by-step solution

The specific heat at constant volume:

The change in internal energy with regard to temperature is what is meant by the definition of the heat capacity at constant volume.

Heat energy at constant volume is given by

$\mathit{Q}\mathbf{=}\mathit{n}{\mathit{C}}_{\mathbf{p}}\mathbf{∆}\mathit{T}$

$\mathit{Q}\mathbf{=}\mathit{n}{\mathit{C}}_{\mathbf{p}}\left(T_b-T_a\right)$ ….. (1)

Here, Q is the amount of heat, n is the number of moles, $C_P$ is the heat capacity at constant pressure, $∆T$is the change in temperature.

(a) Calculate the number of moles using heat energy

Consider the given data as below.

The amount heat, $Q=-2.5\times {10}^{4}J$

Temperature, $T_2=283K$

Temperature, $T_1=323K$

T The heat capacity at constant volume, $C_V=20.76J/kg.K$

The heat capacity at constant pressure, $C_P=29.07J/mol.K$

Rearrange equation (1) for the number of moles as below.

$$\begin{gathered} n=\frac{Q}{C_P∆T} \\ =\frac{Q}{C_P\left(T_2-T_1\right)} \\ =\frac{-2.5\times {10}^4}{29.07\times \left(283-323\right)} \\ =21.5moles \end{gathered}$$

Hence, the number of moles, $n=21.5moles$

Calculate the change in internal energy                  

The change in internal energy is given by

$\mathbf{∆}\mathit{U}\mathbf{=}\mathit{n}{\mathit{C}}_{\mathbf{V}}\mathbf{∆}\mathit{T}$

Here, is the heat capacity at constant volume.

Therefore,

$$\begin{gathered} ∆U=n{C}_V∆T \\ =21.5\times 20.76\times \left(283-323\right) \\ =-17.9KJ \end{gathered}$$

The internal energy decreases as change is negative.

Calculate work using first law of thermodynamics

According to the first law of thermodynamics,

$\mathit{W}\mathbf{=}\mathit{Q}\mathbf{-}\mathbf{∆}\mathit{U}$

Here, is the amount of work done by the gas, Q is the amount of heat, and $\mathbf{∆}\mathit{U}$is the change in internal energy.

As given

The amount of heat, $Q=2.5\times {10}^{4}J$

The change in internal energy. $∆U=-17.9KJ$

Work done is given by

$$\begin{gathered} W=-2.5\times {10}^4-\left(-17.9\times {10}^3\right) \\ =-7.1KJ \end{gathered}$$

Since, the work done is negative, work is done on the gas.

Calculate heat when volume is constant

Now, since the volume is constant, the work done would be zero. According to the first law of thermodynamics

$$\begin{gathered} Q=W+∆U \\ =0+∆U \\ =-17.9KJ \end{gathered}$$

The heat liberated at constant volume is less than at constant pressure.