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Application Of Gas Laws In Various Thermodynamic Processes

This Homework Answers document discusses real-world applications of gas laws. Get it today!

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Application of Gas Laws in Various Thermodynamic Processes1.Given an isothermal process where P1=2.5 mmHgP_1 = 2.5\,\text{mm Hg}P1=2.5mmHg,V1=3.5 LV_1 = 3.5\,\text{L}V1=3.5L, and P2=10 mmHgP_2 = 10\,\text{mm Hg}P2=10mmHg,what is thefinal volume V2V_2V2after the pressure changes?Answer:1.π΄π‘ π‘€π‘’π‘˜π‘›π‘œπ‘€π‘“π‘œπ‘Ÿπ‘‘β„Žπ‘’π‘–π‘ π‘œπ‘‘β„Žπ‘’π‘Ÿπ‘šπ‘–π‘π‘π‘Ÿπ‘œπ‘π‘’π‘ π‘ π‘ƒπ‘‰=π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘π‘ π‘œπ‘ƒ1𝑉1=𝑃2𝑉2(3.5𝐿)(2.5π‘šπ‘šπ»π‘”)=𝑉2(10π‘šπ‘šπ»π‘”)π‘ π‘œπ‘‰2=(3.5𝐿)(2.5π‘šπ‘šπ»π‘”)10π‘šπ‘šπ»π‘”=0.875𝐿2.In a process where the volume is constant, the relationship between pressure and temperatureis given by P∝TP\propto TP∝T. Given that P1=4.1 atmP_1 = 4.1\,\text{atm}P1=4.1atm,T1=100∘CT_1 = 100^\circ CT1=100∘C, and T2=185∘CT_2 = 185^\circ CT2=185∘C, calculate thefinal pressure P2P_2P2.Answer:2.π‘†π‘–π‘›π‘π‘’π‘‘β„Žπ‘’π‘£π‘œπ‘™π‘’π‘šπ‘’π‘–π‘ π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘π‘ π‘œπ‘“π‘œπ‘Ÿπ‘‘β„Žπ‘–π‘ π‘π‘Ÿπ‘œπ‘π‘’π‘ π‘ π‘ƒβˆπ‘‡β„Žπ‘’π‘›π‘π‘’π‘ƒ1𝑇1=𝑃2𝑇2𝑇1=100℃=100+273=373𝐾𝑇2=185℃=185+273=478𝐾𝑃1=4.1π‘Žπ‘‘π‘š4.1373=𝑃2478𝑃2=5.254π‘Žπ‘‘π‘š3.Given that the pressure is constant, use Charles' Law to calculate the final temperature T2T_2T2.Given:β€’Initial volume V1=25 LV_1 = 25\,\text{L}V1=25Lβ€’Initial temperature T1=10∘CT_1 = 10^\circ CT1=10∘Cβ€’Final volume V2=50 LV_2 = 50\,\text{L}V2=50LAnswer:

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