What is the final volume of a 400.0 mL gas sample that is subjected to a

temperature change from 22.0 °C to 30.0 °C and a pressure change from

1000 mmHg to 2 atm? (Watch your pressure!)

Answer: 0.27L

Explanation:

Given that,

Original volume V1 = 400.0 mL

convert volume in milliliters to liters

(If 1000mL = 1L

400.0 mL = 400.0/1000 = 0.4 L)

Original temperature T1 = 22.0 °C

Convert temperature in Celsius to Kelvin

(22.0 °C + 273 = 295K)

Original pressure = 1000mmHg

Convert pressure of 1000mmHg to atm

(If 760mmHg = 1 atm

1000mmHg = 1000/760 = 1.316 atm)

New volume V2 = ?

New Temperature T2 = 30.0°C

(30.0°C + 273 = 303K)

New pressure P2 = 2 atm

Since pressure, volume and temperature are involved, apply the general gas equation

(P1V1) T1 = (P2V2) / T2

(1.316 atm x 0.4 L) / 295K = (2 atm x V2) / 303K

0.526 atmL / 295K = 2V2 / 303K

Cross multiply

0.526 atmL x 303K = 2V2 x 295K

159.47 = 590V2

Divide both sides by 590

159.47/590 = 590V2/590

0.27 L = V2

Thus, the final volume of the gas is 0.27L