Question

At 35.0°C and 3.00 atm pressure, a gas has a volume of 1.40 L. What pressure does the gas have at 0.00°C and a volume of 0.950 L? Which equation should you use? A balloon containing 0.500 mol Ar at 0.00°C and 65.0 kPa pressure is expanded by adding more argon. How many moles of argon are added to bring the sample to a final volume of 60.0 L at 30.0°C and 45.0 kPa? What is the original volume of the gas? L

Answer 1

Answer :

The final pressure of gas will be, 3.92 atm

The original volume of gas is, 17.46 L

The number of moles of argon gas added is, 0.57 mol.

Explanation :

Part 1 :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 3.00 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 1.40 L

$V_2$ = final volume of gas = 0.950 L

$T_1$ = initial temperature of gas = $35.0^oC=273+35.0=308K$

$T_2$ = final temperature of gas = $0.00^oC=273+0.00=273K$

Now put all the given values in the above equation, we get:

$\frac{3.00atm\times 1.40L}{308K}=\frac{P_2\times 0.950L}{273K}$

$P_2=3.92atm$

Therefore, the final pressure of gas will be, 3.92 atm

Part 2 :

First we have to calculate the original volume of gas.

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 65.0 kPa

V = volume of gas = 3.06 L

T = temperature of gas = $0.00^oC=273+0.00=273K$

n = number of moles of gas = 0.500 mol

R = gas constant   = 8.314 kPa.L/mol.K

Now put all the given values in the ideal gas equation, we get:

$(65.0kPa)\times V=(0.500mol)\times (8.314kPa.L/mol.K)\times (273K)$

$V=17.46L$

Now we have to calculate the final moles of sample of gas.

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 45.0 kPa

V = volume of gas = 60.0 L

T = temperature of gas = $30.0^oC=273+30.0=303K$

n = number of moles of gas = ?

R = gas constant   = 8.314 kPa.L/mol.K

Now put all the given values in the ideal gas equation, we get:

$(45.0kPa)\times (60.0L)=n\times (8.314kPa.L/mol.K)\times (303K)$

$n=1.07mol$

Now we have to calculate the number of moles of argon gas added.

Moles of argon gas added = Final moles of gas - Initial moles of gas

Moles of argon gas added = 1.07 mol - 0.500 mol

Moles of argon gas added = 0.57 mol

Thus, the number of moles of argon gas added is, 0.57 mol.