Question

1. gas has a pressure of 1.50atm and volume of 25.00ml. If the volume is decreased to 12.50ml what is the new pressure?

2. 238L ballon is filled with 88.0g of C02 at 15°C , what is the pressure of the gass?​

Answer 1

Given :

•Pressure of the gas =1.50 atm

•volume of the gas=25.00ml

•Volume of the gas is decreased =12.50ml

To Find :

• The new pressure of the gas =?

Formula :

$\\ \red{\underline{\bf{Boyle's \: Law}}}$

$\\ \blue\bigstar{\boxed{\sf{P_{1}V_{1}=P_{2}V_{2}}}}$

Solution :

$\green{\bold{Values}} \begin{cases} \sf{P_{1}=1.50 atm } \\ \\ \sf{P_{2}=?} \\ \\ \sf{V_{1}=25.00ml }\\ \\ \sf{V_{2}=12.50ml} \end{cases}$

By using formula :

P1 V 1=P2 V2

Substituting the given values :

$\\ \\ \implies\sf{P_{1}V_{1}=P_{2}V_{2}}$

$\\ \\ \implies\sf{1.50 \times 25 =P_{2} \times 12.50}$

$\\ \\ \implies\sf{P_{2}=\dfrac{1.50 \times 25}{12.50}}$

$\\ \\ \implies\sf{P_{2}=\dfrac{37.5}{12.50}}$

$\\ \\ \implies\boxed{\sf{P_{2}=3atm}}$

New pressure of the gas is 3 atm .

________________________________________

Given :

•Volume of the gas =238L

•weight (g) =88.0g of co2

•Temperature of the gas =15 degree Celsius

To Find :

• Pressure of the gas =?

Formula :

$\\ \underline{\red{\bf{By \: ideal \: law}}}$

$\\ \blue\bigstar\boxed{\sf{PV=nRT}}$

Solution :

$\red{\bold{Values}} \begin{cases} \sf{Pressure=?} \\ \\ \sf{Volume \: of \: the \: gas =238L} \\ \\ \sf{Weight=88.0g \: of \: CO_{2}} \\ \\ \sf{Temperature=15 \degree C} \end{cases}$

$\\ \\ \implies \sf \: moles \: = \frac{88}{12 + 36} = \frac{88}{48} = \frac{22}{12} = \frac{11}{6} = 1.83$

By using Ideal gas law :

$\\ \implies\sf{PV=nRT}$

$\\ \\ \implies\sf{P \times 238=1.83 \times 0.0821 \times (15 +273)k}$

$\\ \\ \implies\sf{P=\dfrac{1.83 \times 0.0821 \times 288}{238}}$

$\\ \\ \implies\sf{P=\dfrac{43.269}{238}}$

$\\ \\ \implies\sf{P=0.181 atm}$

Pressure of the gas is 0.181 atm.