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The temperature of a sample of gas in a steel


tank at 30.0 kPa is increased from -100.0°C to


25.0 °C. What is the final pressure inside the


tank?


A. 5.17 kPa


B. 51.7 kPa


C. 517 kPa


D. 5170 kPa

Respuesta :

sebassandin

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.

aabdulsamir

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

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