Helium gas in a balloon occupies 2.5 L at 300.0 K. The balloon is dipped into liquid nitrogen that is at a temperature of 77.0 K. What will be the volume of the helium in the balloon at the lower temperature?

A. 0.642 L
B. 1.560 L
C. 9.740 L
D. 9240.0 L

Respuesta :

Answer:

A. 0.641 L

Explanation:

Given data:

V1 = 2.5 L

T1 = 300.0 K

T2= 77.0 K

To find:

V2= ?

Formula :

By using Charles’ Law

[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]

[tex]\frac{V_1}{T_1} . {T_2} = {V_2}[/tex]

Calculation:

V2 = 2.5 L x 77.0 K / 300.0 K

    = 192.5 / 300.0 k

V2    = 0.641 L