Question

A gas has a pressure of 699.0 mm Hg at 40.0°C. What is the temperature at standard pressure (1 atm = 760 mm Hg)?

a) 48.75°C
b) 99.0°C
c) 33.75°C
d) 40.0°C

Answer 1

Using Gay-Lussac's Law and converting Celsius to Kelvin, the temperature at standard pressure when given the temperature and pressure at a different state is calculated. After applying the law and performing calculations, the temperature at standard pressure is determined to be 33.75°C.

Step-by-step explanation:

To find the temperature at standard pressure when given the temperature at a different pressure, we can use Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its temperature (in Kelvin) when the volume and the amount of gas are held constant. The equation for this law can be written as P1/T1 = P2/T2, where P is the pressure and T is the temperature in Kelvin.

First, we convert 40.0°C to Kelvin by adding 273.15 (40.0 + 273.15 = 313.15 K). Then, we apply Gay-Lussac's Law:

(699.0 mm Hg / 313.15 K) = (760 mm Hg / T2)

By cross-multiplication and solving for T2, we get T2= (760 mm Hg * 313.15 K) / 699.0 mm Hg.

Performing the calculation gives us T2 in Kelvin, which we then convert back to Celsius by subtracting 273.15. The final answer is the temperature at standard pressure in Celsius, which is option (c) 33.75°C.