A sample of oxygen at 32◦C occupies 340 mL.
If this sample later occupies 1064 mL at 57◦C
and 1.2 atm, what was its original pressure?
P1*V1/T1=P2*V2/T2
P1= P2*V2*T1/(T2*V2)
change temps to kelvins, then get your calculator out.
To find the original pressure of the oxygen sample, we can use the combined gas law equation, which is as follows:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = original pressure
V1 = original volume
T1 = original temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
Given:
V1 = 340 mL
T1 = 32°C = 32 + 273 = 305 K
V2 = 1064 mL
T2 = 57°C = 57 + 273 = 330 K
P2 = 1.2 atm
Now, we can substitute the values into the combined gas law equation:
(P1 * 340 mL) / (305 K) = (1.2 atm * 1064 mL) / (330 K)
To isolate P1, we can cross multiply:
(P1 * 340 mL * 330 K) = (1.2 atm * 1064 mL * 305 K)
Next, we can divide both sides by (340 mL * 330 K):
P1 = (1.2 atm * 1064 mL * 305 K) / (340 mL * 330 K)
Now we can simplify:
P1 = (1.2 * 1064 * 305) / 340
P1 = 122.784
Therefore, the original pressure of the oxygen sample was approximately 122.784 atm.