A 12.7 L sample of gas is under a pressure of 740 mm Hg at 20degrees C.

What will be the volume of the gas if the pressure increases to 1.00 atm and the temperature drops to 0 degrees C

Use (P1V1)/T1 = (P2V2)/T2

Don't forget to change T to Kelvin.

Use the combined gas law:

P2V2/T2=P1V1/T1 temps in Kelvins

To solve this problem, we can use the combined gas law, which relates the initial and final conditions of pressure, volume, and temperature.

The combined gas law equation is as follows:
(P1 * V1) / (T1) = (P2 * V2) / (T2)

Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature

Given:
P1 = 740 mm Hg
V1 = 12.7 L
T1 = 20 degrees C
P2 = 1.00 atm
T2 = 0 degrees C

First, we need to convert the given values to the appropriate units. We can convert the initial pressure to atm and the initial temperature to Kelvin.

Converting pressure:
1 atm = 760 mm Hg
So, P1 = 740 mm Hg / 760 mm Hg/atm = 0.974 atm

Converting temperature to Kelvin:
T(K) = T(°C) + 273.15
T1 = 20°C + 273.15 = 293.15 K
T2 = 0°C + 273.15 = 273.15 K

Now we can plug the values into the combined gas law equation and solve for V2:

(0.974 atm * 12.7 L) / (293.15 K) = (1.00 atm * V2) / (273.15 K)

Cross-multiplying and solving for V2:
0.974 atm * 12.7 L * 273.15 K = 1.00 atm * V2 * 293.15 K
V2 = (0.974 atm * 12.7 L * 273.15 K) / (1.00 atm * 293.15 K)
V2 ≈ 11.285 L

Therefore, the volume of the gas will be approximately 11.285 L when the pressure increases to 1.00 atm and the temperature drops to 0°C.