Question

Some nitrogen gas is trapped at a pressure of 4.04 atm and a temperature of 761 C in a space that occupies 36.7 L. What would be the temperature of the gas, in K, change to if the pressure changed to 5.38 atm and the volume changes to 99.8 L?

Answer 1

The temperature of the gas would change to 1399.23 K if the pressure changed to 5.38 atm and the volume changed to 99.8 L.

Step-by-step explanation:

We can use the combined gas law to solve this problem:

(P1 x V1) / T1 = (P2 x V2) / T2

where P1, V1, and T1 are the initial pressure, volume, and temperature, and P2, V2, and T2 are the final pressure, volume, and temperature.

We are given:

P1 = 4.04 atm

V1 = 36.7 L

T1 = 761 C = 1034.15 K (convert to Kelvin)

P2 = 5.38 atm

V2 = 99.8 L

Substituting these values into the equation, we get:

(4.04 atm x 36.7 L) / 1034.15 K = (5.38 atm x 99.8 L) / T2

Simplifying, we get:

T2 = (5.38 atm x 99.8 L x 1034.15 K) / (4.04 atm x 36.7 L)

T2 = 1399.23 K

The temperature of the gas would change to 1399.23 K if the pressure changed to 5.38 atm and the volume changed to 99.8 L.