Question

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A 2.0 L balloon contains gas at 300.0 K. The balloon is warmed until the volume is 3.0 L. What is the new temperature of the gas in the balloon?
T = k

Answer 1

450k

Step-by-step explanation:

V1 = 2.0L

T1 = 300.0k

V2 = 3.0L

T2 = ?

V1/T1 = V2/T2 (Charles law)

2.0/300.0 = 3.0/T2

CROSS MULTIPLY

2.0×T2 = 3.0×300.0

DIVIDE BOTH SIDES BY 2.0

2.0×T2/2.0 = 3.0×300.0/2.0

T2 = 900/2.0

= 450K

Answer 2

Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K and 1 atm (101.325 kPa).