1. A gas has a pressure of 799.0 mm Hg at 50.0 degrees C. What is the temperature at standard Pressure?

2. A gas has a pressure of 0.470 atm at 60.0 degrees C. What is the pressure at standard temperature?

A = 674.33mmHg

B = 0.385atm

Explanation:

Both question A and B requires the application of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided that volume is kept constant.

Mathematically,

P = kT, k = P / T

P1 / T1 = P2 / T2 = P3 / T3 == Pn/Tn

A)

Data:

P1 = 799mmHg

T1 = 50°C = (50 + 273.15) = 323.15K

P2 = ?

T2 = 273.15K

P1 / T1 = P2 / T2

Solve for P2

P2 = (P1 × T2) / T1

P2 = (799 × 273.15) / 323.15

P2 = 674.37mmHg

The final pressure is 674.37mmHg

B)

P1 = 0.470atm

T1 = 60°C = (60 + 273.15)K = 333.15K

P2 = ?

T2 = 273.15K

P1 / T1 = P2 / T2

Solve for P2,

P2 = (P1 × T2) / T1

P2 = (0.470 × 273.15) / 333.15

P2 = 0.385atm

The final pressure is 0.385atm

1.

2.

Explanation:

Hello,

1. In this case, the standard pressure is 1 atm or 760 mmHg, therefore, using the Gay-Lussac's law we can compute the corresponding temperature considering the initial 50.0 °C in absolute Kelvins:

2. As well as in the previous case, we now compute the pressure at 273 K which is the standard temperature:

Best regards.

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )