Question

Volume of a gas at STP, if its volume is 80.0 mL at 1.08 atm and –12.5oC.

Answer 1

0.09L

Step-by-step explanation:

Step 1:

Data obtained from the question.

Volume (V) = 80 mL

Pressure (P) = 1.08 atm

Temperature (T) = –12.5°C

Step 2:

Conversion to appropriate unit.

We must convert the unit of the volume and temperature to the right unit of measurement this is illustrated below:

For volume:

1000mL = 1 L

Therefore, 80 mL = 80/1000 = 0.08L

For temperature:

Temperature (Kelvin) = temperature (celsius) + 273

Temperature (Kelvin) = –12.5°C + 273

Temperature (Kelvin) = 260.5K

Step 3:

Determination of the number of mole of the gas.

The number of mole of the gas can be obtained as follow:

Volume (V) = 0.08L

Pressure (P) = 1.08 atm

Temperature (T) = 260.5K

Gas constant (R) = 0.082atm.L/Kmol

Number of mole (n) =?

PV = nRT

1.08 x 0.08 = n x 0.082 x 260.5

Divide both side by 0.082 x 260.5

n = (1.08 x 0.08) / (0.082 x 260.5)

n = 4.04x10^-3 mole

Step 4:

Determination of the volume occupied by the gas at stp.

This is illustrated below:

1 mole of a gas occupy 22.4L at stp

Therefore, 4.04x10^-3 mole will occupy = 0.09L at stp.

Answer 2

90.5mL is the volume of the gas at STP

Step-by-step explanation:

It is possible to find volume of a gas when conditions of temperature and pressure change using combined gas law:

`(P_1V_1)/(T_1) =(P_2V_2)/(T_2)`

Where P is pressure, V is volume and T is absolute temperature. 1 is initial conditions and 2 final conditions.

If initial conditions are 1.08atm, 80.0mL and absolute temperature is (-12.5°C + 273.15) = 260.65K.

And STP are 1atm of pressure and 273.15K of absolute temperature. Replacing:

`(1.08atm80.0mL)/(260.65K) =(1atmV_2)/(273.15K)`

V₂ = 90.5mL is the volume of the gas at STP