Question

A sample of gas initially occupies 3.35 L at a pressure of 0.950 atm at 13.0oC. What will the volume (in L) be if the temperature is changed to 22.5oC, and the pressure is changed to 1.05 atm?

Please answer ASAP

Answer 1

Answer: V= 3.13 L

Explanation: solution attached:

Use combine gas law equation:

P1 V1 / T1 = P2 V2/ T2

Derive to find V2

V2 = P1 V1 T2 / T1 P2

Convert temperatures in K

T1= 13.0°C + 273 = 286 K

T2= 22.5°C + 273 = 295.5 K

Substitute the values.

Answer 2

The branch of science which deals with chemicals and bonds is called chemistry.

The correct answer is Answer 3.13 L

An ideal gas is defined as one for which both the volume of molecules and forces between the molecules are so small that they have no effect on the behavior of the gas.

Numerically the ideal gas is as follows:-

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

All the value is given in the question and we have to find the volume of the second liquid.

The conversion of temperatures in K is as follows:-

$T1= 13.0^oC + 273 = 286 KT2= 22.5^oC + 273 = 295.5 K$

Substitute the values, we will get.

$V_2 =\frac{P_1V_1T_1}{T_2P_2} \\\\V_2 = \frac{0.950*3.35*295.5}{280*1.15} \\\\V_2 =3.13L$

Hence, the correct answer is 3.13L.

For more information, refer to the link:-

https://brainly.com/question/19524691