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

We are given the initial conditions of the air sample and asked to find the minimum pressure required for compression at a constant temperature. This scenario can be solved using Boyle's Law, which states that for a fixed amount of gas at constant temperature, the pressure and volume are inversely proportional.

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 is constant at 30°C.

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:

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

Substituting 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 to 200 dm³ at 30°C is 2.5 bar.

This problem involves a change in the volume of a gas while the temperature and the amount of gas remain constant. We can use Boyle's Law to determine the new pressure.

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 is constant at 35°C.

According to Boyle's Law, for a fixed amount of gas at constant temperature:

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

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

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

Substituting the given values:

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

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

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

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

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 $pV\; =\; nRT$, 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 absolute temperature of the gas

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

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

Substituting this into the ideal gas equation:

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

Now, let's rearrange this equation to relate it to density ($d$). Density is defined as mass per unit volume ($d\; =\; \backslash frac\{m\}\{V\}$). We can rearrange the equation to group $\backslash frac\{m\}\{V\}$:

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

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

$$pM\; =\; dRT$$

To show that density is proportional to pressure at a given temperature, we can rearrange this equation to solve for density:

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

At a given temperature ($T$), the molar mass ($M$) of a specific gas is constant, and $R$ is also a constant. Therefore, for a given gas at a constant temperature:

$\backslash frac\{M\}\{RT\}\; =\; \backslash text\{constant\}$

This implies that:

$d\; \backslash propto\; p$

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