Question

A fluorescent light tube has an internal volume of 400 cm and an internal pressure of 200 kPa.

It is filled with 0.03 moles of an ideal gas.

What is the temperature of the gas inside the fluorescent light tube?

A 3.21 x 10 K

B 3.21 x 10?K

с 3.21 x 105K

D 3.21 x 10K
with explanation plz​

Answer 1

Temperature of gas is 3.21×10² K

Step-by-step explanation:

Given data:

Volume of bulb = 400 cm³ (400/1000 = 0.4L)

Pressure of bulb = 200 KPa (200/101= 1.974 atm)

Number of moles of gas = 0.03 mol

Temperature of gas = ?

Solution:

Formula:

PV = nRT

R = general gas constant = 0.0821 atm.L/ mol.K

1.974 atm × 0.4 L= 0.03 mol × 0.0821 atm.L/ mol.K × T

0.7896 atm.L = 0.002463 atm.L/K × T

T = 0.7896 atm.L / 0.002463 atm.L/K

T = 321 K 0r 3.21×10² K