At 0.00 °C and 1.00 atm pressure, a sample of gas occupies 20.0 mL. If the temperature is increased to 30.0°C and the entire gas sample is transferred to a 30.0-mL container. The pressure of the gas in the new container is:
A 0.499 atm
B 0.740 atm
C 1.35 atm
D 1.66 atm
To solve this problem, we can use the combined gas law equation:
(P1)(V1) / (T1) = (P2)(V2) / (T2)
Where:
P1 = initial pressure = 1.00 atm
V1 = initial volume = 20.0 mL
T1 = initial temperature = 0.00 °C = 273.15 K
P2 = final pressure (what we are trying to solve for)
V2 = final volume = 30.0 mL
T2 = final temperature = 30.0 °C = 303.15 K
Plugging in the values:
(1.00 atm)(20.0 mL) / (273.15 K) = (P2)(30.0 mL) / (303.15 K)
Solving for P2:
(20.0) / (273.15) = (P2)(30.0) / (303.15)
0.0733 = 0.099 P2
P2 = 0.0733 / 0.099
P2 = 0.740 atm
Therefore, the pressure of the gas in the new container is 0.740 atm. So, the correct answer is B) 0.740 atm.