Solveeit Logo

Question

Question: The final volume (in L) of one mole of an ideal gas initially at \({27^ \circ }C\) and 8.21atmpressu...

The final volume (in L) of one mole of an ideal gas initially at 27C{27^ \circ }C and 8.21atmpressure, if it absorbs 420 cal of heat during a reversible isothermal expansion is?

Explanation

Solution

First calculate the initial volume using ideal gas equation PV = nRT
Then we can use that value in the formula for reversible isothermal expansion for final volume.

Complete step-by-step solution:
The formula for work done in a reversible isothermal expansion of an ideal gas is,

W = nRT\ln \dfrac{{{V_{\left( {final} \right)}}}}{{{V_{\left( {initial} \right)}}}} W = 2.303nRT\log \dfrac{{{V_{\left( {final} \right)}}}}{{{V_{\left( {initial} \right)}}}} \\\ \end{gathered} $$(Or) So, let us first find out the value of initial volume using the ideal gas equation, PV=nRT. Given the initial pressure is 8.21atm, Initial temperature is ${27^ \circ }C$ which upon converting to Kelvin scale is 27 + 273K = 300K Gas constant R (in cal) is 2 and the number of moles, n is one. Considering Boyle's law, we have P$ \propto $V which implies the formula for work done in Reversible isothermal expansion in the following manner, $\begin{gathered} W = {\kern 1pt} {\kern 1pt} 2.303nRT\log \dfrac{{{P_{\left( {initial} \right)}}}}{{{P_{\left( {final} \right)}}}} 420 = {\kern 1pt} 2.303 \times {\kern 1pt} 2 \times {\kern 1pt} 300{\kern 1pt} \log \dfrac{{8.21}}{{{P_{\left( {final} \right)}}}} \log {\kern 1pt} 8.21 - log{P_{\left( {final} \right)}} = \dfrac{{420}}{{2.303 \times 600}} \\\ \log {P_{\left( {final} \right)}} = 0.9143 - {\kern 1pt} 0.304 {P_{\left( {final} \right)}} = {\kern 1pt} 4.07atm \end{gathered} $ Now we have final pressure 4.07atm, temperature being isothermal (constant), temperature remains 300k even after expansion. R is 0.0821. To get the final volume let us substitute the values in the ideal gas equation. PV=nRT $\begin{gathered} 4.07 \times V{\kern 1pt} = {\kern 1pt} 1 \times 0.0821{\kern 1pt} \times 300 V = \dfrac{{24.63}}{{4.07}} {V_{\left( {final} \right)}} = 6L \end{gathered} $ **So, the amount of gas which has expanded isothermally is 6litres.** **Additional information:** Dalton’s law is one of the ideal gas laws in which relation between pressure and volume is given. Where PV=constant which implies pressure and volume are inversely related. **Note:** Remember that the heat absorbed has to be taken in positive (+ve) notation. Because we know that the absorption of heat energy is an endothermic reaction.