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Anonymous
Anonymous asked in Science & MathematicsChemistry · 8 years ago

Combined Gas Law Problems, please help me. I'm acutally so lost.?

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?

1 Answer

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  • 8 years ago
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    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

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