Answer:
3.55atm
Explanation:
We will apply Boyle's law formula in solving this problem.
P1V1 = P2V2
And with values given in the question
P1=initial pressure of gas = 1.75atm
V1=initial volume of gas =7.5L
P2=final pressure of gas inside new piston in atm
V2=final volume of gas = 3.7L
We need to find the final pressure
From the equation, P1V1 = P2V2,
We make P2 subject
P2 = (P1V1) / V2
P2 = (1.75×7.5)/3.7
P2=3.55atm
Therefore, the new pressure inside the piston is 3.55atm
Answer:
The answer is either A or C figure it out sorry
<u>Given:</u>
Mass of ice = mass of water = 5.50 kg = 5500 g
Temperature of ice = -20 C
Temperature of water = 75 C
<u>To determine:</u>
Mass of propane required
<u>Explanation:</u>
Heat required to change from ice to water under the specified conditions is:-
q = q(-20 C to 0 C) + q(fusion) + q (0 C to 75 C)
= m*c(ice)*ΔT(ice) + m*ΔHfusion + m*c(water)*ΔT(water)
= 5500[2.10(0-(-20)) + 334 + 4.18(75-0)] = 3792 kJ
The enthalpy change for the combustion of propane is -2220 kJ/mol
Therefore, the number of moles of propane corresponding to the required energy of 3792 kJ = 1 mole * 3792 kJ/2220 kJ = 1.708 moles of propane
Molar mass of propane = 44 g/mol
Mass of propane required = 1.708 moles * 44 g/mol = 75.15 g
Ans: 75.15 grams of propane must be combusted.
To solve this we assume that the gas inside the
balloon is an ideal gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant temperature and number of moles of the gas
the product of PV is equal to some constant. At another set of condition of
temperature, the constant is still the same. Calculations are as follows:
P1V1 = P2V2
V2 = P1V1 / P2
V2 = 0.865 x 1.25 / 0.820
V2 = 1.32 L
