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Application Of Gas Laws In Various Thermodynamic Processes

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    Application of Gas Laws in Various Thermodynamic Processes 1. Given an isothermal process where P1=2.5 mm HgP_1 = 2.5 \ , \ text{mm Hg}P1 =2.5mm Hg, V1=3.5 LV_1 = 3.5 \ , \ text{L}V1 =3.5L, and P2=10 mm HgP_2 = 10 \ , \ text{mm Hg}P2 =10mm Hg, what is the final volume V2V_2V2 after the pressure changes? Answer: 1. 𝐴𝑠 𝑀𝑒 π‘˜π‘›π‘œπ‘€ π‘“π‘œπ‘Ÿ 𝑑 β„Ž 𝑒 π‘–π‘ π‘œπ‘‘ β„Ž π‘’π‘Ÿπ‘šπ‘–π‘ π‘π‘Ÿπ‘œπ‘π‘’π‘ π‘  𝑃𝑉 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ π‘ π‘œ 𝑃 1 𝑉 1 = 𝑃 2 𝑉 2 ( 3 . 5 𝐿 ) ( 2 . 5 π‘šπ‘š 𝐻𝑔 ) = 𝑉 2 ( 10 π‘šπ‘š 𝐻𝑔 ) π‘ π‘œ 𝑉 2 = ( 3 . 5 𝐿 ) ( 2 . 5 π‘šπ‘š 𝐻𝑔 ) 10 π‘šπ‘š 𝐻𝑔 = 0 . 875 𝐿 2. In a process where the volume is constant, the relationship between pressure and temperature is 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 the final pressure P2P_2P2 . Answer: 2. 𝑆𝑖𝑛𝑐𝑒 𝑑 β„Ž 𝑒 π‘£π‘œπ‘™π‘’π‘šπ‘’ 𝑖𝑠 π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ π‘ π‘œ π‘“π‘œπ‘Ÿ 𝑑 β„Ž 𝑖𝑠 π‘π‘Ÿπ‘œπ‘π‘’π‘ π‘  𝑃 ∝ 𝑇 β„Ž 𝑒𝑛𝑐𝑒 𝑃 1 𝑇 1 = 𝑃 2 𝑇 2 𝑇 1 = 100℃ = 100 + 273 = 373 𝐾 𝑇 2 = 185℃ = 185 + 273 = 478 𝐾 𝑃 1 = 4 . 1 π‘Žπ‘‘π‘š 4 . 1 373 = 𝑃 2 478 𝑃 2 = 5 . 254 π‘Žπ‘‘π‘š 3. 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 =25L β€’ Initial temperature T1=10 ∘ CT_1 = 10^ \ circ CT1 =10 ∘ C β€’ Final volume V2=50 LV_2 = 50 \ , \ text{L}V2 =50L Answer:

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    3. 𝐺𝑖𝑣𝑒𝑛 𝑑 β„Ž π‘Žπ‘‘ π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ 𝑖𝑠 π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ π‘ π‘œ 𝑒𝑠𝑖𝑛𝑔 𝑐 β„Ž π‘Žπ‘Ÿπ‘™π‘’π‘  π‘™π‘Žπ‘€ 𝑀𝑒 β„Ž π‘Žπ‘£π‘’ 𝑉 𝑇 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ β„Ž π‘’π‘Ÿπ‘’ 𝑉 1 = 25 𝐿 𝑇 1 = 10℃ = 10 + 273 = 283 𝐾 𝑉 2 = 50 𝐿 𝑇 2 = ? 𝑉 1 𝑇 1 = 𝑉 2 𝑇 2 25 283 = 50 𝑇 2 π‘ π‘œ 𝑇 2 = 566 𝐾 π‘œπ‘Ÿ 566 βˆ’ 273 = 293℃ 4. Using the ideal gas equation, calculate the number of moles (nnn) of a gas. Given the following conditions: β€’ Pressure P=16 torrP = 16 \ , \ text{torr}P=16torr β€’ Volume V=12 LV = 12 \ , \ text{L}V=12L β€’ Temperature T=253 KT = 253 \ , \ text{K}T=253K β€’ Ideal gas constant R=0.0821 L β‹… atm/K β‹… molR = 0.0821 \ , \ text{L} \ cdot \ text{atm} / \ text{K} \ cdot \ text{mol}R=0.0821L β‹… atm/K β‹… mol Answer: 4. 𝑒𝑠𝑖𝑛𝑔 π‘–π‘‘π‘’π‘Žπ‘™ π‘”π‘Žπ‘  π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 𝑀𝑒 β„Ž π‘Žπ‘£π‘’ 𝑃𝑉 𝑅𝑇 = 𝑛 𝑀 β„Ž π‘’π‘Ÿπ‘’ 𝑛 𝑖𝑠 π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘”π‘Žπ‘  P = 16 torr = 16(0.0013)atm =0.0208 atm [1 torr = 0.0013 atm] R =0.0821 L.atm/K mol V =12L T = 253K So 𝑛 = 𝑃𝑉 𝑅𝑇 = ( 0 . 0208 ) ( 12 ) ( 0 . 0821 ) ( 253 ) = 0 . 012 π‘šπ‘œπ‘™ 5. G iven the following conditions for an ideal gas: β€’ Pressure P=2 atmP = 2 \ , \ text{atm}P=2atm β€’ Volume V=6.7 LV = 6.7 \ , \ text{L}V=6.7L

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    β€’ Temperature T=245 KT = 245 \ , \ text{K}T=245K β€’ Number of moles n=0.666 moln = 0.666 \ , \ text{mol}n=0.666mol β€’ Weight of the gas = 34 g Using the ideal gas equation, calculate the molar mass of the gas. Answer: 5. 𝑒𝑠𝑖𝑛𝑔 π‘–π‘‘π‘’π‘Žπ‘™ π‘”π‘Žπ‘  π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 𝑀𝑒 β„Ž π‘Žπ‘£π‘’ 𝑃𝑉 𝑅𝑇 = 𝑛 𝑀 β„Ž π‘’π‘Ÿπ‘’ 𝑛 𝑖𝑠 π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘”π‘Žπ‘  𝑝 = 2 π‘Žπ‘‘π‘š 𝑉 = 6 . 7 𝐿 𝑇 = 245 𝐾 𝑛 = 2 ( 6 . 7 ) ( 245 ) ( 0 . 0821 ) = 0 . 666 π‘šπ‘œπ‘™ 𝑔𝑖𝑣𝑒𝑛 𝑑 β„Ž π‘Žπ‘‘ 𝑀𝑒𝑖𝑔 β„Ž 𝑑 π‘œπ‘“ 𝑑 β„Ž 𝑒 π‘”π‘Žπ‘  = 34 π‘”π‘š 𝑀𝑒𝑖𝑔 β„Ž 𝑑 π‘œπ‘“ 0 . 666 π‘šπ‘œπ‘™ π‘œπ‘“ π‘”π‘Žπ‘  = 34 π‘”π‘š β„Ž 𝑒𝑛𝑐𝑒 𝑀𝑒𝑖𝑔 β„Ž 𝑑 π‘œπ‘“ 1 π‘šπ‘œπ‘™π‘’ π‘”π‘Žπ‘  = 34 0 . 666 = 51 π‘”π‘š 6. Given the following conditions for an ideal gas undergoing a change with constant moles: β€’ Initial pressure P1=277 mm Hg=277(0.0013)=0.3606 atmP_1 = 277 \ , \ text{mm Hg} = 277(0.0013) = 0.3606 \ , \ text{atm}P1 =277mm Hg=277(0.0013)=0.3606atm β€’ Initial volume V1=400 mL=0.4 LV_1 = 400 \ , \ text{mL} = 0.4 \ , \ text{L}V1 =400mL=0.4L β€’ Initial temperature T1=65 ∘ C=65+273=338 KT_1 = 65^ \ circ C = 65 + 273 = 338 \ , \ text{K}T1 =65 ∘ C=65+273=338K β€’ Final pressure P2=1.4 atmP_2 = 1.4 \ , \ text{atm}P2 =1.4atm β€’ Final temperature T2=100 ∘ C=100+273=373 KT_2 = 100^ \ circ C = 100 + 273 = 373 \ , \ text{K}T2 =100 ∘ C=100+273=373K Answer:

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