1. ⇒ (MHT CET 2023 11th May Evening Shift )

If $N_{2}$ gas is compressed at 2 atmosphere from 9.0 L to $3.0 L$ at $300 K$ , find the final pressure at same temperature.

A. 1.66 atm

B. 3.32 atm

C. 6.0 atm

D. 9.0 atm

Correct Option is (C)

According to Boyle's law,

$P_{1} V_{1} = P_{2} V_{2}$ (at constant $n$ and $T$ )

$∴ P_{2} = \frac{P_{1} V_{1}}{V_{2}} = \frac{2 \times 9.0}{3.0} = 6.0 bar$

At constant temperature, the volume of a given amount of gas is inversely proportional to its pressure. Therefore, if the volume is reduced to one-third of its original value, the pressure will correspondingly increase by a factor of three. Hence, correct answer is option (C).

2. ⇒ (MHT CET 2021 20th September Evening Shift )

Keeping temperature constant the pressure of 11.2 dm $^{3}$ of a gas was increased from 105 kPa to 420 kPa. What is the new volume of gas?

A. 1.4 dm $^{3}$

B. 7.0 dm $^{3}$

C. 5.6 dm $^{3}$

D. 2.8 dm $^{3}$

Correct Option is (D)

$\begin{array}{ll} V_{1} = 11.2 {dm}^{3}, & P_{1} = 105 kPa \\ V_{2} = ?, & P_{2} = 420 kPa \end{array}$

According to Boyle's law,

$\begin{aligned} P_{1} V_{1} = P_{2} V_{2} \\ ∴ & V_{2} = \frac{P_{1} V_{1}}{P_{2}} = \frac{105 kPa \times 11.2 {dm}^{3}}{420 kPa} = 2.8 {dm}^{3} \end{aligned}$

3. ⇒ (MHT CET 2021 20th September Morning Shift )

A gas occupies 11.2 dm $^{3}$ at 105 kPa. What is it's volume if pressure is increased to 210 kPa?

A. 22.4 dm $^{3}$

B. 33.6 dm $^{3}$

C. 5.6 dm $^{3}$

D. 16.8 dm $^{3}$

Correct Option is (C)

$\begin{aligned} P_{1} = 105 kPa, V_{1} = 11.2 {dm}^{3} \\ P_{2} = 210 kPa, V_{2} = ? \end{aligned}$

According to Boyle's law, $P_{1} V_{1} = P_{2} V_{2}$

$∴ V_{2} = \frac{P_{1} V_{1}}{P_{2}} = \frac{105 kPa \times 11.2 {dm}^{3}}{210 kPa} = 5.6 {dm}^{3}$

⇒States of Matter