Question

A gas having volume of 10 dm3 is enclosed in vessel at 0°C and
pressure is 2.5 atm. This gas is allowed to expand until the new
pressure is 3 atm. What will be the new volume?​

Answer 1

Given :

• Temperature of vessel = 0° C
• Initial volume ( V₁ ) of gas = 10 dm³
• Initial pressure ( P₁ ) of gas = 2.5 atm
• New pressure ( P₂ ) of gas = 3 atm

Note - We are not provided any information about new temperature , so we will assume that temperature doesn't changed during whole process, that is ,

• Initial temperature = New temperature = 0°C

To find :

New volume ( V₂ ) of gas

Solution :

Using Boyle's law at constant temperature pressure of gas is inversely proportional to its volume . That is, P ∝ (1/V)

This gives,

$\sf \dfrac{P_1}{P_2} = \dfrac{V_2}{V_1}$

Therefore on putting values in above equation we get ;

$\sf \implies \: \dfrac{2.5 \: atm}{3 \: atm} = \dfrac{V_2}{10 \: {dm}^{3} }$

$\sf \implies \: V_2 = \: \dfrac{2.5 \: }{3 \: } \times \: 10 \: {dm}^{3}$

$\sf \implies \: V_2 = \: \dfrac{25 \: }{3 \: } \: {dm}^{3}$

$\sf \implies \: V_2 = \bold{\: 8.33 \: {dm}^{3} }$

ANSWER : 8.33 dm³