Home > Physics > Gas Laws

A container with a volume of #25 L# contains a gas with a temperature of #150^o K#. If the temperature of the gas changes to #320 ^o K# without any change in pressure, what must the container's new volume be?

Answer 1

Layla Ahmed

#160/3L#

From Ideal Gas Equation; #PV=nRT#

Pressure, number of moles and Gas constant are all fixed.

#V=nRT/P# or #V/T=nR/P#

As long as pressure and the amount of gas remains unchanged, the ratio of #V# and #T# is always the same.

#V_1/T_1=V_2/T_2# #(25L)/(150K)=(V_2)/(320K)# #V_2=160/3L#

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

Answer 2

Carson Alexander

Using the combined gas law, ( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} ), and given that pressure remains constant:

( \frac{V_2}{V_1} = \frac{T_2}{T_1} )

Plugging in the values:

( \frac{V_2}{25, \text{L}} = \frac{320, \text{K}}{150, \text{K}} )

( V_2 ≈ \frac{320, \text{K}}{150, \text{K}} \times 25, \text{L} )

( V_2 ≈ 53.33, \text{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.