Answer:
[tex]P_2=51.7kPa[/tex]
Explanation:
Hello,
In this case, we apply the Gay-Lussac's law which allows us to understand the pressure-temperature behavior via a directly proportional relationship:
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
Thus, since we are asked to compute the final pressure we solve for it in the previous formula, considering the temperature in absolute Kelvin units:
[tex]P_2=\frac{P_1T_2}{T_1}=\frac{30.0kPa*(25.0+273)K}{(-100.0+273)K} \\\\P_2=51.7kPa[/tex]
Best regards.
Answer:
B = Pressure = 51.7kPa
Explanation:
P1 = 30kPa
T1 = -100°C = (-100 + 273.15)K = 173.15K
T2 = 25°C = (25 + 273.15)K = 298.15K
P2 = ?
This question involves the use of pressure (p) law which states that the pressure of a fixed mass of gas is directly proportional to its temperature(t) provided that the volume of the gas is kept constant.
Mathematically,
P = kT, k = P / T
P1 / T1 = P2 / T2 = P3 / T3 =.......= Pn / Tn
P1 / T1 = P2 / T2
Solve for P2,
P2 = (P1 × T2) / T1
P2 = (30 × 298.15) / 173.15
P2 = 8944.5 / 173.15
P2 = 51.66kPa
The pressure of the gas is approximately 51.7kPa
