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

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:

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:

Best regards.

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

the ratio of the conjugate base to the acid required to attain the desired ph.