A container with a volume of #48 L# contains a gas with a temperature of #140^o K#. If the temperature of the gas changes to #190 ^o K# without any change in pressure, what must the container's new volume be?

Question

A container with a volume of #48 L# contains a gas with a temperature of #140^o   K#.  If the temperature of the gas changes to #190     ^o    K# without any change in pressure, what must the container's new volume be?

Nora Arzate · Accepted Answer

#65.143\ L#

Assuming the gas to be ideal then using ideal gas equation then by Charles law

At constant pressure, for a closed system

#V\propT#

#\frac{V_1}{V_2}=\frac{T_1}{T_2}#

#V_2=V_1(T_2/T_1)#

Given that the initial volume of gas is #V_1=48\ L# at initial temperature #T_1=140 \ K#. Now the final new volume #V_2# at temperature #T_2=190\ K# keeping pressure constant

#V_2=48(190/140)#

#=456/7#

#=65.143\ L#

Isabella Alvarez · Answer

undefined To find the new volume of the gas, we can use the formula for Charles's law:

V1 / T1 = V2 / T2

Where:
V1 = Initial volume of the gas (48 L)
T1 = Initial temperature of the gas (140 K)
V2 = Final volume of the gas (unknown)
T2 = Final temperature of the gas (190 K)

Rearranging the formula to solve for V2, we get:

V2 = (V1 * T2) / T1

Substituting the values, we get:

V2 = (48 L * 190 K) / 140 K
V2 = (48 * 190) / 140
V2 = 9120 / 140
V2 = 65.14 L

Therefore, the container's new volume should be approximately 65.14 liters.