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 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑖𝑐 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑠𝑜
𝑃1𝑉1 = 𝑃2𝑉2
(3.5𝐿)(2.5𝑚𝑚 𝐻𝑔) = 𝑉2(10𝑚𝑚 𝐻𝑔)
𝑠𝑜 𝑉2 = (3.5𝐿)(2.5 𝑚𝑚 𝐻𝑔)
10𝑚𝑚 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 𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑠𝑜 𝑓𝑜𝑟 𝑡ℎ𝑖𝑠 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑃 ∝ 𝑇
ℎ𝑒𝑛𝑐𝑒 𝑃1
𝑇1
= 𝑃2
𝑇2
𝑇1 = 100℃ = 100 + 273 = 373𝐾
𝑇2 = 185℃ = 185 + 273 = 478𝐾
𝑃1 = 4.1 𝑎𝑡𝑚
4.1
373 = 𝑃2
478
𝑃2 = 5.254 𝑎𝑡𝑚
3. 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=50L
Answer:
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 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑖𝑐 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑠𝑜
𝑃1𝑉1 = 𝑃2𝑉2
(3.5𝐿)(2.5𝑚𝑚 𝐻𝑔) = 𝑉2(10𝑚𝑚 𝐻𝑔)
𝑠𝑜 𝑉2 = (3.5𝐿)(2.5 𝑚𝑚 𝐻𝑔)
10𝑚𝑚 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 𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑠𝑜 𝑓𝑜𝑟 𝑡ℎ𝑖𝑠 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑃 ∝ 𝑇
ℎ𝑒𝑛𝑐𝑒 𝑃1
𝑇1
= 𝑃2
𝑇2
𝑇1 = 100℃ = 100 + 273 = 373𝐾
𝑇2 = 185℃ = 185 + 273 = 478𝐾
𝑃1 = 4.1 𝑎𝑡𝑚
4.1
373 = 𝑃2
478
𝑃2 = 5.254 𝑎𝑡𝑚
3. 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=50L
Answer:
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Subject
Chemistry