5
Use the ideal gas law: P1/T1 = P2/T2. Rearrange the equation to solve for P2: P2 = (P1/T1) * T2. Plug in the values: P2 = (325 kPa / 283 K) * 60 degrees Celsius. Convert the temperature to Kelvin: 60 degrees Celsius + 273 = 333 K. Calculate the new pressure: P2 ≈ 361 kPa.
Yes. 85 psi is equal to about 586.1 kPa
Assuming the amount of gas remains constant, we can use the ideal gas law to calculate the final absolute pressure. The initial pressure (P1) is 200 kPa and the final volume (V2) is 250 cm3. The initial temperature (T1) is 40 degrees Celsius or 313.15 Kelvin, and the final temperature (T2) is 20 degrees Celsius or 293.15 Kelvin. Using the equation (P1 * V1) / T1 = (P2 * V2) / T2, we can solve for the final absolute pressure (P2), which is approximately 400 kPa.
(psi x 6.89476 = kPa). So, 70 psi x 6.89476 = 482.633 kPa
783,0 mm Hg is equal to 104,3914 kPa.
0.6 kPa
The vapor pressure of kerosene at 25 degrees Celsius is approximately 0.28 kPa.
The vapor pressure of water at 105 degrees Celsius is approximately 101.3 kilopascals (kPa).
Water will boil at approximately 120.6 degrees Celsius when the external pressure is 31 kPa. This is lower than the standard boiling point of water at 100 degrees Celsius due to the reduced pressure.
Use the ideal gas law: P1/T1 = P2/T2. Rearrange the equation to solve for P2: P2 = (P1/T1) * T2. Plug in the values: P2 = (325 kPa / 283 K) * 60 degrees Celsius. Convert the temperature to Kelvin: 60 degrees Celsius + 273 = 333 K. Calculate the new pressure: P2 ≈ 361 kPa.
My mistake, its should be about 111 degrees celsius.
The boiling point of ethanoic acid (acetic acid) at 65 kPa is approximately 104 degrees Celsius.
The vapor pressure of water at 70 degrees Celsius is approximately 23.76 kPa. To find the partial pressure of water vapor in the mixture, subtract this vapor pressure from the total pressure of 89.9 kPa. Therefore, the partial pressure of water vapor would be 89.9 kPa - 23.76 kPa = 66.14 kPa.
how many moles are contained in 4.67 L sample of gas at 33 degrees celcius and 199 kpa
See the Web Links to the left for a table of the vapor pressure of water at various temperatures.
1 atm=760 mm Hg= 760 torr =101.325 kPa and 0 degrees Celsius
Using the ideal gas law (P1/T1 = P2/T2), we can set up the equation as: 125 kPa / (30.0 + 273.15 K) = 201 kPa / T2. Solving for T2, the new temperature would be approximately 57.7 degrees Celsius.