A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to 1.3 atm, and the temperature of the gas increases to 285 K.
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A very simple question just apply ideal gas equation....
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Hey dear,
◆ Answer-
V2 = 231.1 cm^3
◆ Explaination-
# Given-
V1 = 168 cm^3
T1 = 255 K
P1 = 1.6 atm
T2 = 285 K
P2 = 1.3 atm
# Solution-
Using ideal gas eqn,
PV = nRT
As no of moles remains constant,
n = P1.V1/R.T1 = P2.V2/R.T2
V2 = P1.V1.T2 / P2.T1
V2 = (1.6 × 168 × 285) / (1.3 × 255)
V2 = 231.1 cm^3
Therefore, final volume of gas is 231.4 cm^3.
Hope it helps...
◆ Answer-
V2 = 231.1 cm^3
◆ Explaination-
# Given-
V1 = 168 cm^3
T1 = 255 K
P1 = 1.6 atm
T2 = 285 K
P2 = 1.3 atm
# Solution-
Using ideal gas eqn,
PV = nRT
As no of moles remains constant,
n = P1.V1/R.T1 = P2.V2/R.T2
V2 = P1.V1.T2 / P2.T1
V2 = (1.6 × 168 × 285) / (1.3 × 255)
V2 = 231.1 cm^3
Therefore, final volume of gas is 231.4 cm^3.
Hope it helps...
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