Problem 10.44 - Enhanced with Feedback T; T: where P is pressure, V is volume, a
ID: 696586 • Letter: P
Question
Problem 10.44 - Enhanced with Feedback T; T: where P is pressure, V is volume, and T' is temperature can be utilized to calculate the desired volume. When rearranged to solve for V2 it becomes Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 55.0 gallons and which contains O2 gas at a pressure of 16,500 kPa at 21°GC PiViT2 (165000 kPa)(208.20 L) (273 K) (296 K)(101.325 kPal-3.15 x 104 L You may want to reference (pages 404-407) Section 10.4 while completing this problem Part C At what temperature would the pressure in the tank equal 140.0 atm? Express the answer in atmospheres to three significant figures T- 3.82.104 Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining Part D What would be the pressure of the gas, in kPa, if it were transferred to a container at 24 °C whose volume is 55.0 L? Express the answer in kilopascals to three significant figures kPa Submit My Answers Give UpExplanation / Answer
gas law equation can be written as
P1V1/T1= P2V2/T2
where P1, V1, T1 , P2.V2 and T2 are pressure, volume, temperature at conditions 1 and 2 respectively..
when V1= V2
P1/T1= P2/T2
T2= P2*T1/P1
given P2 =140 atm, P1= 16500Kpa, 101.3 Kpa= 1atm, 16500 Kpa= 16500/101.3 atm =162.88 atm
T1 = 21 deg.c= 21+273= 294K,
T2= `140*294/162.88 K=252.7K
2. for the second case, volume remained constant
hence P1/T1= P2/T2
P2= P1*T2/T1= 16500*(24+273)/(21+273)= 16668.37 Kpa