Answer to Question #193633 in Chemistry for Valentina Rojas

Question

What is the temperature (in degrees Celsius) of 6.02 mol of hydrogen sulfate gas in a 100 L container at 0.95 atm?

Answer 1

P = 0.95 atm

V = 100 L

n = 6.02 mol

R = 0.0821 atm L mol^-1 K^-1

T = unknown

Solution:

The Ideal Gas equation can be used.

The Ideal Gas equation can be expressed as: PV = nRT

To find the temperature, solve the equation for T:

T = PV / nR

T = (0.95 atm × 100 L) / (6.02 mol × 0.0821 atm L mol^-1 K^-1) = 192.2 K

T (K) = t (°C) + 273.15

Hrnce,

t = T - 273.15 = 192.2 - 273.15 = -80.95°C = -81°C

t = -81°C

Answer: -81°C