CBSE Class 11 Chemistry Chapter 5: States of Matter NCERT Solutions

We are given the initial conditions of the air sample and the desired final volume. Since the temperature is stated to remain constant (30°C), we can apply Boyle's Law, which describes the inverse relationship between the pressure and volume of a gas at constant temperature.

Given:

• Initial pressure, $p\_1\; =\; 1$ bar
• Initial volume, $V\_1\; =\; 500\; \backslash text\{\; dm\}^3$
• Final volume, $V\_2\; =\; 200\; \backslash text\{\; dm\}^3$
• Temperature, T = constant

According to Boyle's Law:

$$p\_1\; V\_1\; =\; p\_2\; V\_2$$

We need to find the final pressure, $p\_2$. Rearranging the formula to solve for $p\_2$:

$$p\_2\; =\; \backslash frac\{p\_1\; V\_1\}\{V\_2\}$$

Now, substitute the given values:

$$p\_2\; =\; \backslash frac\{(1\; \backslash text\{\; bar\})\; \backslash times\; (500\; \backslash text\{\; dm\}^3)\}\{200\; \backslash text\{\; dm\}^3\}$$

$$p\_2\; =\; \backslash frac\{500\}\{200\}\; \backslash text\{\; bar\}$$

$$p\_2\; =\; 2.5\; \backslash text\{\; bar\}$$

Therefore, the minimum pressure required to compress the air is 2.5 bar.

This problem involves the transfer of a gas from one container to another at a constant temperature. This scenario is governed by Boyle's Law, which states that for a fixed amount of gas at constant temperature, the product of pressure and volume remains constant.

Given:

• Initial pressure, $p\_1\; =\; 1.2$ bar
• Initial volume, $V\_1\; =\; 120\; \backslash text\{\; mL\}$
• Final volume, $V\_2\; =\; 180\; \backslash text\{\; mL\}$
• Temperature, T = 35 °C (constant)

According to Boyle's Law:

$$p\_1\; V\_1\; =\; p\_2\; V\_2$$

We need to find the final pressure, $p\_2$. Rearranging the formula:

$$p\_2\; =\; \backslash frac\{p\_1\; V\_1\}\{V\_2\}$$

Substitute the given values into the equation:

$$p\_2\; =\; \backslash frac\{(1.2\; \backslash text\{\; bar\})\; \backslash times\; (120\; \backslash text\{\; mL\})\}\{180\; \backslash text\{\; mL\}\}$$

Calculate the result:

$$p\_2\; =\; \backslash frac\{144\}\{180\}\; \backslash text\{\; bar\}$$

$$p\_2\; =\; 0.8\; \backslash text\{\; bar\}$$

Therefore, the pressure of the gas in the larger vessel would be 0.8 bar.

The equation of state for an ideal gas is given by:

$$pV\; =\; nRT$$ .....(i)

Where:

• $p$ is the pressure of the gas
• $V$ is the volume of the gas
• $n$ is the number of moles of the gas
• $R$ is the ideal gas constant
• $T$ is the temperature of the gas (in Kelvin)

We know that the number of moles ($n$) can be expressed in terms of the mass of the gas ($m$) and its molar mass ($M$) as:

$$n\; =\; \backslash frac\{m\}\{M\}$$

Substitute this expression for $n$ into the ideal gas equation (i):

$$pV\; =\; \backslash left(\backslash frac\{m\}\{M\}\backslash right)RT$$

Now, let's rearrange this equation to isolate the term $\backslash frac\{m\}\{V\}$, which represents the density ($d$) of the gas:

$$pM\; =\; \backslash frac\{m\}\{V\}RT$$

Since density $d\; =\; \backslash frac\{m\}\{V\}$, we can substitute $d$ into the equation:

$$pM\; =\; dRT$$

To show that density is proportional to pressure at a given temperature, we rearrange the equation to solve for $d$:

$$d\; =\; \backslash frac\{pM\}\{RT\}$$

At a given temperature ($T$), the molar mass of the gas ($M$) is constant, and the ideal gas constant ($R$) is also a constant. Therefore, the terms $\backslash frac\{M\}\{RT\}$ are constant.

This implies that the density ($d$) is directly proportional to the pressure ($p$):

$$d\; \backslash propto\; p\; \backslash quad\; (\backslash text\{at\; constant\; T\})$$

Thus, we have shown that at a given temperature, the density of a gas is directly proportional to its pressure.