Question

A sample of cl2 gas has a volume of 15.0 l at a pressure of 1.50 atm and a temperature of 23 °c. What volume, in liters, will the gas occupy at 3.50 atm and 286 °c? Assume ideal behavior.

Answer 1

Volume of Cl₂ gas at 3.5 atm and 286°C = 12.14 L

Further explanation

Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated), among others  

• Boyle's law at constant T, P = 1 / V  

• Charles's law, at constant P, V = T  

• Avogadro's law, at constant P and T, V = n  

So that the three laws can be combined into a single gas equation, the ideal gas equation  

In general, the gas equation can be written  

$\large{\boxed{\bold{PV=nRT}}}$

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

R = gas constant = 0.082 l.atm / mol K  

T = temperature, Kelvin

State 1

V1= 15 L

P1 = 1.5 atm

T1 = 23°C +273 = 296 K

We can find mole(n)

$\displaystyle n=\frac{PV}{RT}\\\\n=\frac{1.5.15}{0.082.296}\\\\n=0.927$

State 2

P2=3.5 atm

T = 286°C +273 = 559 K

n (from state 1) = 0.927

$\displaystyle V=\frac{n.R.T}{P}\\\\V=\frac{0.927.0.082.559}{3.5}\\\\V=\boxed{\bold{12.14~L}}$

Learn more  

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Keywords : ideal gas law, Cl₂

Answer 2

Given: Initial volume of $CO_{2}$ gas= 15.0 L

Pressure=1.50 atm

Temperature=23°C

Use ideal gas equation as follows:

$PV=nRT$

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature in Kelvin.

First convert the given temperature in Kelvin.

0°C=273.15 K

Thus, 23 °C=296.15 K

Rearranging ideal gas equation to calculate the value of nR as follows:

$nR=\frac{PV}{T}=\frac{(1.50 atm)(15.0 L)}{(296.15 K)}=0.0760 atm LK^{-1}$

Now, final pressure is 3.50 atm and temperature is 286°C, converting the temperature into Kelvin,

0°C=273.15 K

Thus, 286 °C=559.15 K

Volume can be calculated using the ideal gas equation as follows:

$V=\frac{nRT}{P}$

Putting the value of given pressure and temperature and the above calculated value of nR,

$V=\frac{(0.0760 atmLK^{-1})(559.15 K)}{(3.50 atm)}=12.14 L$

Therefore, volume of $CO_{2}$ at 3.50 atm and 286°C is 12.14 L.