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Sagot :
We have a gas that changes its volume, pressure and temperature. We will assume that there is no entry or exit of matter from the container. Therefore the moles will remain constant.
We have two states, an initial state (1) and a final state (2) for the gas. If we assume that the gas behaves as an ideal gas we can apply the following equation:
[tex]PV=nRT[/tex]Where,
P is the pressure of the gas, in atm
V is the volume of the gas, in liters
n is the number of moles of the gas
T is the temperature of the gas, in Kelvin
R is a constant, 0.08206 atm.L/mol.K
For each state we will have that the equation will be:
Initial state (1)
[tex]\begin{gathered} P_1V_1=nRT_1 \\ nR=\frac{P_1V_1}{T_1} \end{gathered}[/tex]P1 = 100.0kPa = 0.987atm
V1 = 2.66L
T1 = 39.1°C = 312.25K
Final state (2)
[tex]\begin{gathered} P_2V_2=nRT_2 \\ nR=\frac{P_2V_2}{T_2} \end{gathered}[/tex]P2 = 544.0kPa = 5.369atm
T2= 62.2°C = 335.35K
V2 = Unknown
Since R and n are constant and equal in both states of the gas, we can equate the equations and we have:
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]We clear V2 and replace the known data:
[tex]V_2=\frac{P_1V_1}{T_1}\times\frac{T_2}{P_2}[/tex][tex]\begin{gathered} V_2=\frac{0.987atm\times2.66L}{312.25K}\times\frac{335.35K}{5.369atm} \\ V_2=0.525L \end{gathered}[/tex]Answer: The new volume in liters will be 0.525
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