Anonymous
Anonymous asked in Science & MathematicsChemistry · 8 years ago

2. A gas has a volume of 39 L at STP. What will its volume be at 4 atm and 25°C?

3. A gas, now contained at STP, has a volume of 500 mL. The initial pressure was 0.96 atm at 20°C. What was the initial volume?

5.

Calculate the pressure in a tire if it starts out filled with 7.54 L of air at 219 kPa and 21.6°C and gets heated to 65.2°C as the volume increases to 7890 mL.

8.

6.8 L of a gas is found to exert 97.3 kPa at 25.2°C. If the volume is changed to 12.5 L, what would be the required temperature to change the pressure to standard pressure?

Relevance
• 8 years ago

2.)

At STP, the pressure will be 1 atm. So you have to use the combined gas law:

P1V1/T1=P2V2/T2

Where:

T1=273K

T2=25+273=298K

P1=1 atm

P2=4 atm

V1= 39 L

V2=?

V2=V1*(P1/P2)*(T2/T1)

V2=39L*(1atm/4atm)*(298K/273K)

V2=10.6L

3.)

At STP, the pressure will be 1 atm. So you have to use the combined gas law as well:

P1V1/T1=P2V2/T2

Where:

T1=20+273=293K

T2=273K

P1=0.96 atm

P2=1 atm

V1= ?

V2=500mL

V1=V2*(P2/P1)*(T1/T2)

V1=500mL*(1 atm/0.96 atm)*(293K/273K)

V1=559mL

5.)

You will have to use the combined gas law as well; the concepts are the same:

P1V1/T1=P2V2/T2

Where:

T1=21.6+273=294.6K

T2=65+273K=338K

P1=219 kPa

P2=?

V1= 7.54L

V2=7890mL=7.890L

P2=P1*(V1/V2)*(T2/T1)

P2=219kPa*(7.54L/7.890L)*(338K/294.6K)

P2=240 kPa

8.)

You will have to use the combined gas law as well; the concepts are the same:

The pressure of a gas at STP is 1 atm=101.325 kPa

P1V1/T1=P2V2/T2

Where:

T1=25.2+273=298.2K

T2=?

P1=97.3 kPa

P2=101.325 kPa

V1= 6.8L

V2=12.5L

T2=T1*(P2/P1)*(V2/V1)

T2=298.2K*(97.3kPa/101.325kPa)*(6.8L/12.5L)

T2=156K