Question

In Problems 50 through 53 , the initial state of 1.00 mol1.00 \text{~mol} of a dilute gas is P1=3.00 atm,V1=1.00 LP_1=3.00 \text{~atm}, V_1=1.00 \text{~L}, and Eint 1=456 JE_{\text {int } 1}=456 \text{~J}, and its final state is P2=2.00 atm,V2=3.00 LP_2=2.00 \text{~atm}, V_2=3.00 \text{~L}, and Eint 2=912 JE_{\text {int } 2}=912 \text{~J}.

The gas is allowed to expand isothermally until it reaches its final volume and its pressure is 1.00 atm1.00 \text{~atm}. It is then heated at constant volume until it reaches its final pressure. (a)(a) Illustrate this process on a PVP V diagram and calculate the work done by the gas.

Solution

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Consider the given initial and final state data:

n=1.00 molP1=3.00 atm=303.975 kPaV1=1.00 L=1103 m3Eint 1=456 JP2=2.00 atm=202.65 kPaV2=3.00 L=3103 m3Eint 2=912 J\begin{align*} n &=1.00~\text{mol} \\ P_{1} &=3.00 \mathrm{~atm}=303.975~\text{kPa} \\ V_{1} &=1.00 \mathrm{~L}=1\cdot10^{-3}~\text{m}^{3} \\ E_{\text {int } 1} &=456 \mathrm{~J} \\ P_{2} &=2.00 \mathrm{~atm}=202.65~\text{kPa} \\ V_{2} &=3.00 \mathrm{~L}=3\cdot10^{-3}~\text{m}^{3} \\ E_{\text{int } 2} &=912 \mathrm{~J} \end{align*}

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