sample of neon initially has a volume of 4.00 Lat 18 C . Part A What is the new
ID: 508260 • Letter: S
Question
sample of neon initially has a volume of 4.00 Lat 18 C .
Part A
What is the new temperature, in degrees Celsius, when the volume of the sample is changed at constant pressure and amount of gas to 6.00 L ?
T=___
Part B
What is the new temperature, in degrees Celsius, when the volume of the sample is changed at constant pressure and amount of gas to 1400 mL ?
T=______
Part C
What is the new temperature, in degrees Celsius, when the volume of the sample is changed at constant pressure and amount of gas to 6.50 L ?
T=_____
Part D
What is the new temperature, in degrees Celsius, when the volume of the sample is changed at constant pressure and amount of gas to 3800 mL
T=____
sample of neon initially has a volume of 4.00 Lat 18 C .
Part A
What is the new temperature, in degrees Celsius, when the volume of the sample is changed at constant pressure and amount of gas to 6.00 L ?
T=___
Part B
What is the new temperature, in degrees Celsius, when the volume of the sample is changed at constant pressure and amount of gas to 1400 mL ?
T=______
Part C
What is the new temperature, in degrees Celsius, when the volume of the sample is changed at constant pressure and amount of gas to 6.50 L ?
T=_____
Part D
What is the new temperature, in degrees Celsius, when the volume of the sample is changed at constant pressure and amount of gas to 3800 mL
T=____
Explanation / Answer
A)
Find New T if
V1 = 4 L , T1 = 18°C = 291 K
V2 = 6 L, T2 =?
T2 = T1*V2/V1 = 291*4/6 = 194 K = 194 -273 = -79 °C
B)
Find T if we change to V = 1400 mL = 1.4 L
V1 = 4 L , T1 = 18°C = 291 K
V2 = 1.2 L, T2 =?
V1/T1 = V2/T2
T2 = V2/V1*T1 = 4/1.2*291 = 970 K = 970 -273 = 697°C
C)
V1/T1 = V2/T2
4/(291) = 6.5/T2
T2 = 6.5/4*291 = 472 K = 199 °C
D)
V1/T1 = V2/T2
4/(291) = 3.8/T2
T2 = 3.8/4*291 = 276.2 K = 276.2 -273°C = 3.2 °
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