Home > Physics > Gas Laws

A container with a volume of #8 L# contains a gas with a temperature of #370^o C#. If the temperature of the gas changes to #890 ^o K# without any change in pressure, what must the container's new volume be?

Answer 1

Julia Aguilar

The volume is #=11.1L#

Charles' Law is used.

#V_1/T_1=V_2/T_2#

#V_1=8L#

#T_1=370+273=643K#

#T_2=890K#

#V_2=T_2/T_1*V_1#

#=890/643*8=11.1L#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

William Audy

To find the new volume, we can use the formula:

[ V_2 = \frac{{V_1 \times T_2}}{{T_1}} ]

Where:

( V_1 = 8 , \text{L} ) (initial volume)
( T_1 = 370^\circ \text{C} + 273.15 = 643.15 , \text{K} ) (initial temperature in Kelvin)
( T_2 = 890 , \text{K} ) (final temperature in Kelvin)

Substituting the values:

[ V_2 = \frac{{8 \times 890}}{{643.15}} ]

[ V_2 \approx 11.06 , \text{L} ]

So, the container's new volume must be approximately 11.06 L.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.