Question

In a flexible container , 15.9 L of gas is under 589 kPa of pressure at a temperature of 56.5 degrees * C . If the pressure and temperature change to STP , what is the new volume ? A 10.2 B 92.4L 112 D 76.6 L

Answer 1

76.6 litres is the volume of the gas if pressure and temperature changes to STP.

Explanation:

Data given:

Volume of the gas V1 = 15.9 litres

pressure of the gas P1 = 589 kPa or 5.81 atm.

temperature T1 = 56.5 degrees or 329.3 K

Final pressure P2 = 101.3 kPa

Final temperature T2 = 273 K

At STP the pressure is 1 atm or 101.3 kPa and the temperature is 273 K. So from the question the final temperature and pressure is at STP.

from Gas Law:

$\frac{P1V1}{T1}$  =  $\frac{P2V2}{T2}$

V2 = $\frac{P1V1T2}{P2T1}$

Putting the values in  the above rearranged equation of Gas Law:

V2 = $\frac{589 X 15.9 X 273}{329.3X 101.3}$

    = 76.6 Litres

Answer 2

If the pressure and temperature change to STP, the new volume will be "76.6 liters".

Pressure and Temperature

According to the question,

Initial pressure, P₁ = 589 kPa or,

                              = 5.81 atm

Final pressure, P₂ = 101.3 kPa

Initial volume, V₁ = 15.9 L

Initial temperature, T₁ = 56.5° or,

                                    = 329.3 K

Final temperature, T₂ = 273 K

By using the Gas law,

→ $\frac{P_1 V_1}{T_1} =\frac{P_2 V_2}{T_2}$

or, the new volume be:

→ V₂ = $\frac{589\times 15.9\times 273}{329.3\times 101.3}$

       = 76.6 L

Thus the above approach is correct.

Find out more information about Gas law here:

https://brainly.com/question/555495