Question

An ideal gas initially at$\mathbf{300}\mathbf{K}\mathbf{}$is compressed at a constant pressure of$\mathbf{25}\mathbf{}\mathbf{N}\mathbf{/}{\mathbf{m}}^2\mathbf{}$from a volume of$\mathbf{3}\mathbf{.}\mathbf{0}{\mathbf{m}}^3$to a volume of$\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{}{\mathbf{m}}^3\mathbf{}$. In the process,$\mathbf{75}\mathbf{J}\mathbf{}$is lost by the gas as heat. What are

(a) the change in internal energy of the gas and

(b) the final temperature of the gas?

Short answer

1. The internal energy change in the gas is$\Delta E_{\mathrm{int}}=-45\text{J}$ .
2. The temperature of the gas changes to$180\text{\hspace{0.17em}K}$ .

Step-by-step solution

Concept

The process that takes place at constant pressure is called an isobaric process. Using the formulas for work done W in an isobaric process, and the first law of thermodynamics, we can find the change in the internal energy${\mathbf{\Delta E}}_{\mathrm{int}}$of the gas. Also, by using theideal gas law in the ratio form, we can find the final temperature ${\mathbf{T}}_2$of the gas.

1.The work done W in anisobaric process is

$\mathbf{W}\mathbf{=}\mathbf{p\Delta V}$

2.The ideal gas law in ratio form is

${\mathbf{T}}_2\mathbf{=}{\mathbf{T}}_1\left(\frac{{\mathbf{V}}_2}{{\mathbf{V}}_1}\right)$

3.Thechange in the internal energy is

${\mathbf{\Delta E}}_{\mathrm{int}}\mathbf{=}\mathbf{Q}\mathbf{-}\mathbf{W}$

Step 2: Given Data

1. The pressure.$p=25{\text{\hspace{0.17em}N/m}}^{\text{2}}$
2. The final volume$V_f=1.8{\text{m}}^3$
3. The initial volume.$V_i=3.0{\text{m}}^3$
4. The initial temperature$T_1=300\text{K}$
5. In the process, the energy lost by the gas as heat is$Q=-75\text{J}$ .

Step 3: Calculations

a.The work done in anisobaricprocess is expressed as-

$W=p\Delta V$

Where p is the pressure and $\Delta V=(V_f-V_i)$is change in the volume

Therefore,

$\begin{array}{c}W=p(V_f-V_i)\\ =25\text{\hspace{0.17em}N}/{\text{m}}^2\times (1.8{\text{\hspace{0.17em}m}}^3-3.0{\text{\hspace{0.17em}m}}^3)\\ =-30\text{\hspace{0.17em}J}\end{array}$

From the first law of thermodynamics, we have-

$\Delta E_{\mathrm{int}}=Q-W$

Where Q is the heat transferred and$\Delta {\text{E}}_{int}$is the change in the internal energy o the gas. It is given that Q amount of energy leaves the system; therefore, we can consider Q as negative.$Q=-75\text{\hspace{0.17em}J}$

$\begin{array}{c}\Delta E_{\mathrm{int}}=-75\text{\hspace{0.17em}J}-(-30\text{\hspace{0.17em}J})\\ =-75\text{\hspace{0.17em}J}+\left(30\text{\hspace{0.17em}J}\right)\\ =-45\text{\hspace{0.17em}J}\end{array}$

Internal energy is decreased by$-45\text{\hspace{0.17em}J}$

b.Since the pressure is constant and the number of moles presumed are constant, the ideal gas law can be written as-

$\begin{array}{c}{T}_2=T_1\left(\frac{V_2}{V_1}\right)\\ =300\text{K}\times \left(\frac{1.8{\text{m}}^3}{3.0{\text{m}}^3}\right)\\ =180\text{\hspace{0.17em}K}\end{array}$

Conclusion

The change in internal energy of the gas is $-45J$and the final temperature of the gas is $180\text{\hspace{0.17em}K}$.