A 250.mL sample of gas at 1.00 atm and 20.0°C has the temperature increased to 40.0°C and the volume increased to 500.mL.What is the new pressure?

Question

Quizplus · Accepted Answer

Using the combined gas law:

(P1V1)/T1 = (P2V2)/T2

P1 = 1.00 atm (Given)
V1 = 250 mL (Given)
T1 = 20.0°C + 273.15 K = 293.15 K (Conversion from °C to K)
V2 = 500 mL (Given)
T2 = 40.0°C + 273.15 K = 313.15 K (Conversion from °C to K)

Substituting these values into the equation:

(1.00 atm x 250mL) / 293.15K = (P2 x 500mL) / 313.15K

P2 = (1.00 atm x 250mL x 313.15K) / (293.15K x 500mL) = 0.534 atm

Therefore, the new pressure is 0.534 atm, which corresponds to choice C.

A 250ML Sample of Gas at 1