Question

The gas inside a cylinder expands against a constant external pressure of 1.00 atm from a volume of 5.00 L to a volume of 13.00 L. In doing so, it turns a paddle immersed in 1.00 L water. Calculate the temperature increase of the water, assuming no loss of heat to the surroundings or frictional losses in the mechanism.

Short answer

Hence, the change in temperature is 0.09 K.

Step-by-step solution

The given Information.

1. The external pressure is 1 atm.
2. The initial volume is 5.00 L.
3. The final volume is 13.00 L.
4. The volume of water is 1.00 L.

First law of thermodynamics.

It states that if heat is provided to the system, then it changes the internal energy of the system, and the remaining show work done.

$\mathrm{Q}=\mathrm{\Delta U}+\mathrm{w}$

Q is the heat given to the system.

$\mathrm{\Delta U}$is the change in the internal energy.

W is the work done by the system.

Since, the work is done against external pressure, the process is irreversible.

The temperature difference.

The ${\mathrm{P}}_{\mathrm{ext}}$is 1.00 atm.

The gas expands from 5 L to 13 L.

The work done can be found as;

$\begin{array}{c}\mathrm{W}={\mathrm{P}}_{\mathrm{ext}}\mathrm{\Delta V}\\ =1\mathrm{atm}\times \left(13.00\mathrm{L}-5.00\mathrm{L}\right)\\ =8\mathrm{Latm}\end{array}$

1 atm is equal to 101.325 J.

The work done in joules is;

$\begin{array}{c}\mathrm{W}=-8\times 101.325\mathrm{J}\\ =-810.6\mathrm{J}\end{array}$

The formula used to calculate the change in internal energy is;

$\mathrm{\Delta U}={\mathrm{nC}}_{\mathrm{V}}\mathrm{\Delta T}$

Assuming that the gas is ideal,

${\mathrm{C}}_{\mathrm{v}}=\frac{3}{2}\mathrm{RT}$

$\begin{array}{c}\mathrm{\Delta U}=\mathrm{n}\times \frac{3}{2}\mathrm{R\Delta T}\\ =\frac{3}{2}\left(\mathrm{nR\Delta T}\right)\\ =\frac{3}{2}\times {\mathrm{P}}_{\mathrm{ext}}\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)\end{array}$

Substituting the values in the equation,

$\begin{array}{c}\mathrm{\Delta U}=\frac{3}{2}\times {\mathrm{P}}_{\mathrm{ext}}\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)\\ =\frac{3}{2}\times 1\times \left(13-5\right)\mathrm{L}\times 101.325\mathrm{J}\\ =1215.9\mathrm{J}\end{array}$

Calculating heat from the first law of thermodynamics;

$\begin{array}{c}\mathrm{Q}=\mathrm{\Delta U}+\mathrm{w}\\ =\left(1015.9-810.6\right)\mathrm{J}\\ =405.3\mathrm{J}\end{array}$

Temperature change calculation:

The mass can be found as:

$\begin{array}{c}\mathrm{Mass}=\mathrm{Density}\times \mathrm{Volume}\\ =1.00{\mathrm{gcm}}^{-3}\times 1\times 1000{\mathrm{cm}}^3\\ =1000\mathrm{g}\end{array}$

The specific heat (s) of water is equal to $4.18{\mathrm{JK}}^{-1}{\mathrm{g}}^{-1}$.

The change in temperature can be given as;

$\mathrm{Q}=\mathrm{mS\Delta T}$

$\begin{array}{c}405.3\mathrm{J}=1000\mathrm{g}\times 4.18{\mathrm{Jg}}^{-1}{\mathrm{K}}^{-1}\times \mathrm{\Delta T}\\ \mathrm{\Delta T}=\frac{405.3\mathrm{J}}{1000\mathrm{g}\times 4.18{\mathrm{Jg}}^{-1}{\mathrm{K}}^{-1}}\\ =0.09\mathrm{K}\end{array}$