Question

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?

Answer 1

Answer:

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

Answer 2

Answer:

1. $T_2=307.23K=34.23\°C$

2. $P_2=0.385atm$

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:

$\frac{T_1}{P_1}= \frac{T_2}{P_2}\\\\T_2=\frac{T_1P_2}{P_1}=\frac{(50+273.15)K*760mmHg}{799mmHg} \\\\T_2=307.23K=34.23\°C$

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

$\frac{P_1}{T_1}= \frac{P_2}{T_2}\\\\P_2=\frac{P_1T_2}{T_1}=\frac{273K*0.47atm}{(60+273)K} \\\\P_2=0.385atm$

Best regards.