Exploring Robert Boyle's Law: Pressure and Volume of CO2

• MHB
• WMDhamnekar
In summary, Robert Boyle's law states that at constant temperature, the pressure of a fixed amount of gas varies inversely with its volume. This can be represented mathematically as $p ∝ \frac{1}{V}$ and $p \times V = k_1$, where $k_1$ is a proportionality constant dependent on the amount of gas, temperature, and units of pressure and volume. The law also states that if the volume of gas at constant temperature expands, the pressure will decrease proportionally. The given table shows the effect of pressure on the volume of CO₂ at 300 K, but there may be a mistake in the values listed. The volume of CO₂ can also be calculated using the ideal
WMDhamnekar
MHB
Robert Boyle's law states that at constant temperature, the pressure of fixed amount ( i-e number of moles n) of gas varies inversely with its volume. Mathematically, it can be written as $p ∝ \frac1V$(at constant T and n) $\Rightarrow p = k_1 \times \frac1V$ where $k_1$is a proportionality constant.

The value of constant $k_1$ depends upon the amount of the gas, temperature of the gas and the units in which p and V are expressed. $p \times V= k_1$

If a fixed amount of gas at constant temperature T occupying volume $V_1$ at pressure $p_1$ undergoes expansion, so that volume becomes $V_2$ and pressure becomes $p_2,$ then according to Boyle’s law : $p_1 \times V_1 = p_2 \times V_2=$ constant $\Rightarrow \frac{p_1}{p_2} = \frac{V_2}{V_1}.$

It should be noted that volume V of the gas doubles, if pressure is halved.

The following table 5.1 gives effect of pressure on volume of 0.09 mol of CO₂ at 300 K. but i didn't understand these calculated values given in the second column. I also didn't understand the meanings of headings given to each column. If any member can explain me how the values in the second column is computed, may answer to this question.

My understanding:

$V= \frac{nRT}{p}\tag {1}$ where n, R, T, p are constants. n stands for number of moles, R is gas constant, T is temperature and p is pressure.

Putting the given values in this equation (1),we get 11.2 liters =$\frac {0.09 mol \times 8,314 J k^{-1} mol^{-1}\times 300 K }{20000 Pa}$

But in the second column, it is $112 \times 10^{-3} m^3= 112$ liters . How is that? Where i am wrong?Can we compute the volume of $CO_2$ in another way? For example, by using this known information that one mole of $CO_2$ molecules features a volume of 22.414 liters at standard T and p. So, 0.09 mol of $CO_2$ features a volume of $0.09 \times 22.414= 2.01726$ liters at STP.

Last edited:
Which table are we talking about?
Otherwise I can only guess why it would show $112\times 10^{-3}\,m^3$.

Either way, your substitution in the ideal gas law (equation 1) has $p=20\,000\,Pa$, which is not standard pressure.
If instead we substitute the STP values $p_0=100\,000\,Pa$ and $T_0=273.15\,K$, we find:
$$\frac {0.09\, mol \times 8.314\, J K^{-1} mol^{-1}\times 273.15\, K }{100\,000\, Pa} = 2.04\, L$$
which agrees with what you found using the molar volume at STP.
Note that the ideal gas law predicts a slightly higher value (1%), which is because a real gas is slightly more cohesive than an ideal gas.

Last edited:
Klaas van Aarsen said:
Which table are we talking about?
Otherwise I can only guess why it would show $112\times 10^{-3}\,m^3$.

Either way, your substitution in the ideal gas law (equation 1) has $p=20\,000\,Pa$, which is not standard pressure.
If instead we substitute the STP values $p_0=100\,000\,Pa$ and $T_0=273.15\,K$, we find:
$$\frac {0.09\, mol \times 8.314\, J K^{-1} mol^{-1}\times 273.15\, K }{100\,000\, Pa} = 2.04\, L$$
which agrees with what you found using the molar volume at STP.
Note that the ideal gas law predicts a slightly higher value (1%), which is because a real gas is slightly more cohesive than an ideal gas.

Hello,

Sorry. I forgot to add the table. Now, i have added it to my question.

It looks as a mistake in the table. I believe the unit in the pressure column should be $10^3\,Pa$. Then the other 3 columns have the correct values.
We can verify with either molar volume or ideal gas law at $T=300\,K$, or we can look it up with an online calculator or table.

1. What is Robert Boyle's Law?

Robert Boyle's Law, also known as Boyle's Law, is a gas law that describes the relationship between the pressure and volume of a gas at a constant temperature. It states that the pressure of a gas is inversely proportional to its volume, meaning that as pressure increases, volume decreases and vice versa.

2. Who was Robert Boyle?

Robert Boyle was a 17th century Irish scientist and philosopher who is considered one of the founders of modern chemistry. He is best known for his work in gas laws, including Boyle's Law, and for his pioneering experiments in the field of chemistry.

3. How is Boyle's Law used in everyday life?

Boyle's Law is used in a variety of everyday applications, including scuba diving, where it explains why air tanks must be filled with compressed air in order to provide enough oxygen for divers at high pressures. It is also used in the production of carbonated beverages, where carbon dioxide gas is compressed to increase its solubility in liquid.

4. What is the mathematical equation for Boyle's Law?

The mathematical equation for Boyle's Law is P1V1 = P2V2, where P1 and V1 represent the initial pressure and volume of a gas, and P2 and V2 represent the final pressure and volume of the gas. This equation can be rearranged to solve for any of the variables, as long as the temperature remains constant.

5. How does temperature affect Boyle's Law?

Boyle's Law only holds true when the temperature remains constant. If the temperature of a gas changes, the relationship between pressure and volume will also change. This is because temperature affects the kinetic energy of gas molecules, which in turn affects their movement and collisions, ultimately impacting the pressure and volume of the gas.

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