# A sample of oxygen occupies 15.0 mL under a pressure of 0.928 atm at 25.0 C. What would be its volume at another time when the pressure was 0.948 atm and the temperature 19.0 C?

Pls help me

## (0.948)V/(273+19) = (0.928)(15)/(273+25)

## To solve this problem, we can use the ideal gas law equation:

PV = nRT

Where:

P = pressure (in atm)

V = volume (in liters)

n = number of moles

R = gas constant (0.0821 L.atm/mol.K)

T = temperature (in Kelvin)

First, we need to convert the given temperatures from Celsius to Kelvin.

T1 = 25.0 °C + 273 = 298 K

T2 = 19.0 °C + 273 = 292 K

Now, we can use the ideal gas law equation for both conditions and set them equal to each other to find the ratio of volumes:

P1V1 = P2V2

Substituting the given values:

(0.928 atm)(15.0 mL) = (0.948 atm)(V2)

Now, we need to convert the initial volume from mL to L:

15.0 mL = 15.0 mL / 1000 = 0.015 L

Simplifying the equation further:

(0.928)(0.015) = (0.948)(V2)

Now, we can solve for V2:

V2 = (0.928)(0.015) / 0.948

V2 ≈ 0.0146 L

Therefore, the volume of oxygen at another time, when the pressure is 0.948 atm and the temperature is 19.0 °C, would be approximately 0.0146 L.

## To solve this problem, we can use the combined gas law equation:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

where:

P1 = initial pressure

V1 = initial volume

T1 = initial temperature

P2 = final pressure

V2 = final volume (what we need to find)

T2 = final temperature

Let's plug in the given values into the equation:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

P1 = 0.928 atm

V1 = 15.0 mL = 15.0 cm³ (since 1 mL = 1 cm³)

T1 = 25.0 °C = 25.0 + 273.15 K (convert to Kelvin)

P2 = 0.948 atm

T2 = 19.0 °C = 19.0 + 273.15 K (convert to Kelvin)

Now, we can rearrange the equation to solve for V2:

V2 = (P1 * V1 * T2) / (P2 * T1)

Substituting the values:

V2 = (0.928 atm * 15.0 cm³ * (19.0 + 273.15 K)) / (0.948 atm * (25.0 + 273.15 K))

Now, we can calculate:

V2 = (13.92 * (292.15)) / (21.32 * (298.15))

V2 ≈ 12.529 cm³

Therefore, the volume of oxygen at the final conditions (0.948 atm and 19.0 °C) would be approximately 12.529 cm³.