Application Of Gas Laws In Various Thermodynamic Processes

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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.𝐴𝑠𝑤𝑒𝑘𝑛𝑜𝑤𝑓𝑜𝑟𝑡ℎ𝑒𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑖𝑐𝑝𝑟𝑜𝑐𝑒𝑠𝑠𝑃𝑉=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠𝑜𝑃1𝑉1=𝑃2𝑉2(3.5𝐿)(2.5𝑚𝑚𝐻𝑔)=𝑉2(10𝑚𝑚𝐻𝑔)𝑠𝑜𝑉2=(3.5𝐿)(2.5𝑚𝑚𝐻𝑔)10𝑚𝑚𝐻𝑔=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.𝑆𝑖𝑛𝑐𝑒𝑡ℎ𝑒𝑣𝑜𝑙𝑢𝑚𝑒𝑖𝑠𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠𝑜𝑓𝑜𝑟𝑡ℎ𝑖𝑠𝑝𝑟𝑜𝑐𝑒𝑠𝑠𝑃∝𝑇ℎ𝑒𝑛𝑐𝑒𝑃1𝑇1=𝑃2𝑇2𝑇1=100℃=100+273=373𝐾𝑇2=185℃=185+273=478𝐾𝑃1=4.1𝑎𝑡𝑚4.1373=𝑃2478𝑃2=5.254𝑎𝑡𝑚3.Given that the pressure is constant, use Charles' Law to calculate the final temperature T2T_2T2.Given:•Initial volume V1=25 LV_1 = 25\,\text{L}V1=25L•Initial temperature T1=10∘CT_1 = 10^\circ CT1=10∘C•Final volume V2=50 LV_2 = 50\,\text{L}V2=50LAnswer:

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3.𝐺𝑖𝑣𝑒𝑛𝑡ℎ𝑎𝑡𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑖𝑠𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠𝑜𝑢𝑠𝑖𝑛𝑔𝑐ℎ𝑎𝑟𝑙𝑒𝑠𝑙𝑎𝑤𝑤𝑒ℎ𝑎𝑣𝑒𝑉𝑇=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡ℎ𝑒𝑟𝑒𝑉1=25𝐿𝑇1=10℃=10+273=283𝐾𝑉2=50𝐿𝑇2=?𝑉1𝑇1=𝑉2𝑇225283=50𝑇2𝑠𝑜𝑇2=566𝐾𝑜𝑟566−273=293℃4.Using the ideal gas equation, calculate the number of moles (nnn) of a gas. Given the followingconditions:•Pressure P=16 torrP = 16\,\text{torr}P=16torr•Volume V=12 LV = 12\,\text{L}V=12L•Temperature T=253 KT = 253\,\text{K}T=253K•Ideal gas constant R=0.0821 L⋅atm/K⋅molR = 0.0821\,\text{L}\cdot\text{atm} /\text{K}\cdot\text{mol}R=0.0821L⋅atm/K⋅molAnswer:4.𝑢𝑠𝑖𝑛𝑔𝑖𝑑𝑒𝑎𝑙𝑔𝑎𝑠𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑤𝑒ℎ𝑎𝑣𝑒𝑃𝑉𝑅𝑇=𝑛𝑤ℎ𝑒𝑟𝑒𝑛𝑖𝑠𝑛𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑚𝑜𝑙𝑒𝑠𝑜𝑓𝑔𝑎𝑠P = 16 torr = 16(0.0013)atm =0.0208 atm[1 torr = 0.0013 atm]R =0.0821 L.atm/K molV =12LT = 253KSo𝑛=𝑃𝑉𝑅𝑇=(0.0208)(12)(0.0821)(253)=0.012𝑚𝑜𝑙5. Given the following conditions for an ideal gas:•Pressure P=2 atmP= 2\,\text{atm}P=2atm•Volume V=6.7 LV = 6.7\,\text{L}V=6.7L

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•Temperature T=245 KT = 245\,\text{K}T=245K•Number of moles n=0.666 moln = 0.666\,\text{mol}n=0.666mol•Weight of the gas = 34 gUsing the ideal gas equation, calculate the molar mass of the gas.Answer:5.𝑢𝑠𝑖𝑛𝑔𝑖𝑑𝑒𝑎𝑙𝑔𝑎𝑠𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑤𝑒ℎ𝑎𝑣𝑒𝑃𝑉𝑅𝑇=𝑛𝑤ℎ𝑒𝑟𝑒𝑛𝑖𝑠𝑛𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑚𝑜𝑙𝑒𝑠𝑜𝑓𝑔𝑎𝑠𝑝=2𝑎𝑡𝑚𝑉=6.7𝐿𝑇=245𝐾𝑛=2(6.7)(245)(0.0821)=0.666𝑚𝑜𝑙𝑔𝑖𝑣𝑒𝑛𝑡ℎ𝑎𝑡𝑤𝑒𝑖𝑔ℎ𝑡𝑜𝑓𝑡ℎ𝑒𝑔𝑎𝑠=34𝑔𝑚𝑤𝑒𝑖𝑔ℎ𝑡𝑜𝑓0.666𝑚𝑜𝑙𝑜𝑓𝑔𝑎𝑠=34𝑔𝑚ℎ𝑒𝑛𝑐𝑒𝑤𝑒𝑖𝑔ℎ𝑡𝑜𝑓1𝑚𝑜𝑙𝑒𝑔𝑎𝑠=340.666=51𝑔𝑚6.Given the following conditions for an ideal gas undergoing a change with constant moles:•Initial pressureP1=277 mmHg=277(0.0013)=0.3606 atmP_1 = 277\,\text{mm Hg} =277(0.0013) = 0.3606\,\text{atm}P1=277mmHg=277(0.0013)=0.3606atm•Initial volume V1=400 mL=0.4 LV_1 = 400\,\text{mL} = 0.4\,\text{L}V1=400mL=0.4L•Initial temperature T1=65∘C=65+273=338KT_1 = 65^\circ C = 65 + 273 = 338\,\text{K}T1=65∘C=65+273=338K•Final pressure P2=1.4 atmP_2 = 1.4\,\text{atm}P2=1.4atm•Final temperature T2=100∘C=100+273=373KT_2 = 100^\circC = 100 + 273 = 373\,\text{K}T2=100∘C=100+273=373KAnswer:

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