A flexible container is put in a deep freeze. Its original volume is 3.00 m3 at 25.0°C. After the container cools, it has shrunk

Question

fiasKO [112]

3 years ago

5

A flexible container is put in a deep freeze. Its original volume is 3.00 m3 at 25.0°C. After the container cools, it has shrunk

to 2.00 m3. Its new temperature in degrees Celsius is

Chemistry

1 answer:

Goryan [66] · Answer 1 · 2022-09-23T14:16:05+03:00

This is an ideal gas problem. The gas inside the balloon is considered ideal. Ideal gas equation is a function pressure, temperature, amount and volume. Note: amount is constant since the balloon ins closed. Pressure is maintained constant since the walls are flexible. Ideal gas equation is: PV=nRT. Put all constant in one side and variables in one.

P/nR=T/V. To find the answer to the question equate the constants of both situation
T1/V1=T2/V2
(25+273.15)/3=(x+273.15)/2
x=-74.38 degC