Question #29789
if the tempeture of 50.0L of a gas at 40.0 degrees C falls by 10.0 degrees C what is the new volume of the gas if the pressure is constant?
Expert's answer
For an ideal gas at constant pressure, the volume is directly proportional to its temperature.
V1/T1 = V2/T2
where
V is volume
T is temperature (K)
V1 = 50.0 L
T1 = 313 K
If the temperature falls by 10.0 C, so T2 will be = 40 - 10 = 30 C or 303 K
Therefore V2 will be
V2 = V1*T2/T1
V2 = 50.0 L * 303 K / 313 K = 48.40 L
Answer: V2 = 48.40 L
V1/T1 = V2/T2
where
V is volume
T is temperature (K)
V1 = 50.0 L
T1 = 313 K
If the temperature falls by 10.0 C, so T2 will be = 40 - 10 = 30 C or 303 K
Therefore V2 will be
V2 = V1*T2/T1
V2 = 50.0 L * 303 K / 313 K = 48.40 L
Answer: V2 = 48.40 L
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