Asample of nitrogen gas occupies 1.55 l at 27.0°c and 1.00 atm. what will the volume be at −100.0°c and the same pressure?

0.89L ( 2 decimal place)

Explanation:

Using Combined Gas law.

( P₁V₁)/T₁ = (P₂V₂) / T₂

P₁ = 1.00 atm

V₁ = 1.55L

T₁ = 27.0 °C = converting to kelvin ( 27 + 273k) = 300k

P₂ = same as P₁ = 1.00 atm

V₂ = ?

T₂ = -100°C = converting to kelvin ( -100 + 273k) = 173k

(1 x 1.55)/300 = (1 x V)/173

1.55/300 = V/173

V / 173 = 0.00516666666

V = 173 x 0.00516666666 = 0.89383333333 ≈ 0.89L ( 2 decimal place)

The volume at -100 °C is 0.894 L

Explanation:

Step 1: Data given

The initial volume = 1.55L

The initial temperature = 27.0 °C = 300 K

The pressure is 1.00 atm and stays constant

The temperature lowers to -100 °C = 173 K

Step 2: Calculate the volume

V1 / T1 = V2 / T2

⇒ with V1 = the initial volume = 1.55L

⇒ with T1 = the initial temperature = 300 K

⇒ with V2 = The new volume = TO BE DETERMINED

⇒ with T2 = the final temperature = 173 K

1.55 / 300 = V2 /173

V2 =  0.894 L

The volume at -100 °C is 0.894 L