A sample of argon has a volume of 6.0 cm 3 and the pressure is 0.87 atm. If the final temperature is 35 degrees celsius, the final volume is 10 cm 3 and the final pressure is 0.52 atm, what was the initial temperature of the argon

Respuesta :

Answer:

= 36.185 °C

Explanation:

Using the combined gas law;

P1V1/T1 = P2V2/T2

In this case;

P1 = 0.87 atm

V1 = 6.0 cm³

T1 = ?

P2 = 0.52 atm

V2 = 10 cm³

T2 = 35 °C + 273 = 308 K

Therefore;

T1 = P1V1T2/P2V2

   = ( 0.87 × 6.0 × 308)/( 0.52 ×10)

   = 309.185 K

Therefore; Initial temperature = 309.185 K - 273 = 36.185 °C