Problem 1: Volume Calculation at Increased Pressure

Given:

• Initial volume $V_1=12.3$ liters
• Initial pressure $P_1=40.0$ mmHg
• Final pressure $P_2=60.0$ mmHg

Using Boyle's Law, which states that $P_1{V}_1=P_2{V}_2$, we can rearrange the formula to solve for the final volume $V_2$:

$$V_2=\frac{P_1{V}_1}{P_2}$$

Substituting in the known values:

$$V_2=\frac{(40.0\mathrm{\text{mmHg}})(12.3\mathrm{\text{L}})}{60.0\mathrm{\text{mmHg}}}$$

Calculating:

$$V_2=\frac{492.0}{60.0}=8.20\mathrm{\text{L}}$$

Final Volume: 8.20 L

Problem 2: Volume Calculation at Increased Pressure

Given:

• Initial volume $V_1=3.60$ liters
• Initial pressure $P_1=1.00$ atm
• Final pressure $P_2=2.50$ atm

Using Boyle's Law:

$$V_2=\frac{P_1{V}_1}{P_2}$$

Substituting in the known values:

$$V_2=\frac{(1.00\mathrm{\text{atm}})(3.60\mathrm{\text{L}})}{2.50\mathrm{\text{atm}}}$$

Calculating:

$$V_2=\frac{3.60}{2.50}=1.44L$$

Final Volume: 1.44 L

Problem 3: Pressure Calculation for Compression

Given:

• Initial volume $V_1=400.0cu.ft.$
• Initial pressure $P_1=1.00atm$
• Final volume $V_2=3.00cu.ft.$

Using Boyle's Law:

$$P_2=\frac{P_1{V}_1}{V_2}$$

Substituting in the known values:

$$P_2=\frac{(1.00atm)(400.0cu.ft.)}{3.00cu.ft}$$

Calculating:

$$P_2=\frac{400}{3}\approx 133.\overline{3}atm$$

Final Pressure: Approximately 133 atm

Problem 4: Volume Calculation at Increased Pressure

Given:

• Initial volume $V_1=1.56L$
• Initial pressure $P_1=1.00atm$
• Final pressure $P_2=3.00atm$

Using Boyle's Law:

$$V_2=\frac{P_1{V}_1}{P_2}$$

Substituting in the known values:

$$V_2=\frac{(1.00atm)(1.56L)}{3.00atm}$$

Calculating:

$$V_2=\frac{1.56}{3}=0.\overline{5}L$$

Final Volume: Approximately 0.520 L

Problem 5: Pressure Calculation with Changed Volume

Given:

• Initial volume $V_{1}=$11.20 liters
• Initial pressure $P_1=0.\overline{8}atm=0.\overline{860}mmHg)$
• Final volume ( V_{2}=15.\overline {0}, L)

Using Boyle's Law:

$$P_2=\frac{P_1*V_l}{V_l}$$ Substituting in the known values:

[ P_{l}= (11.\overline {20}, L)(0.\overline {860}, mmHg)= (x)(15.\overline {0}, L) \ x= (11.\overline {20})(860)/(15)= .642 mmHg \ x= .642 mmHg

Final Pressure: Approximately .642 mmHg

Problem 6: Volume Conversion at Standard Pressure

Given:

• Initial volume ( V_{i}=500 mL)
• Initial pressure( P_i=745 mmHg)

To find final volume at standard pressure ($760mmHg)$:

Using Boyle's Law,

[ (V_f)= (745)(500)/(760)=490 mL \

Final Volume: Approximately490 mL

Problem 7: Conversion of Gas Volume Under Different Pressures

Given: Initial conditions are as follows: Volume ($350mL)$, and initial pressure ($740mmHg)$.

To find final volume under standard pressure ($760mmHg)$:

Using Boyle’s law,

[ (V_f)= (740)(350)/(760)= x mL \

x= (740*350)/760= x mL \

x=343mL \

Final Volume :Approximately343mL

Problem 8: Conversion of Gas Volume Under Different Pressures

Given: Initial conditions are as follows: Volume ($338L)$, and initial pressure ($63atm)$.

To find final volume under standard pressure ($760mmHg)$:

Using Boyle’s law,

[ (V_f)= (63)(338)/(760)= x L \

x= (63*338)/760= x L \

x=28mL \

Final Volume :Approximately28mL

Problem9 :Conversion of Gas Volume Under Different Pressures.

Given : Initial conditions are as follows : Volume(\273mL), and initial pressure(\166kPa).

To find final volume under standard pressure(\101kPa).

Using Boyle’s law,

[ (V_f)= (166)(273)/(101325)= x mL \

x=(166*273)/101325=x mL \

x=42mL \

Final Volume :Approximately42mL

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