Answer: The answer is 68142.4 Pa
Explanation:
Given that the initial properties of the cylindrical tank are :
Volume V1= 0.750m3
Temperature T1= 27C
Pressure P1 =7.5*10^3 Pa= 7500Pa
Final properties of the tank after decrease in volume and increase in temperature :
Volume V2 =0.480m3
Temperature T2 = 157C
Pressure P2 =?
Applying the gas law equation (Charles and Boyle's laws combined)
P1V1/T1 = P2V2/T2
(7500 * 0.750)/27 =( P2 * 0.480)/157
P2 =(7500 * 0.750* 157) / (0.480 *27)
P2 = 883125/12.96
P2 = 68142.4Pa
Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa
Answer:
C4H7OH + 6O2 => 4CO2 + 4H2O
Explanation:
hope This helps :)
Answer:
V₂ = 116126.75 cm³
Explanation:
Given data:
Radius of balloon = 15 cm
Initial pressure = 2 atm
Initial temperature = 35 °C (35 +273 = 308K)
Final temperature = -20°C (-20+273 = 253 K)
Final pressure = 0.3 atm
Final volume = ?
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
Initial volume of balloon:
V = 4/3πr³
V = 4/3×22/7×(15cm)³
V = 14137.17 cm³
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2 atm × 14137.17 cm³ × 253 K / 308 K × 0.3 atm
V₂ = 7153408.02 atm .cm³. K / 61.6 K.atm
V₂ = 116126.75 cm³
