Answer: 0.08K
Explanation:
When temperature changes, the corresponding change in thermal energy of a gas is given by:
ΔE (thermal) = 3/2nRΔT
Defining the parameters:
ΔE (thermal) = Increase in thermal energy of the mono atomic gas = 1.0J
n = number of moles of the gas = 1.0mol
R = Ideal gas constant = 8.314J/mol/K
ΔT = change in temperature. This is what we need to find.
Rearranging the equation to make ΔT subject of the formula,
ΔT= 2 x ΔE (thermal) / (3 x n x R)
Therefore, ΔT = 2 x 1.0J / (3 x 1.0mol x 8.314J/mol/K)
ΔT = 2.0J / 24.942J/K
ΔT = 0.0802K
ΔT = 0.08K
The temperature change of 1.0mol of a monoatomic gas if its thermal energy is increased by 1.0J is 0.08K.
Answer:
heat is not absorbed by the system
A rapid shift occurs between gas compression and expansion.
All heat is transformed to work done by the system.
Explanation:
Explanation:
The unit of volume derived from SI units(1) is the cubic meter, m3.
Chemists in a laboratory usually deal with much smaller volumes than cubic meters and the metric but non-SI units of liter or litre (L) and milliliter or millilitre (mL or ml) are in common use.
In 1964 the litre was redefined as being equal to exactly 1 cubic decimetre:
1 L = 1 dm3
So 1 milliltre = 1 cubic centimetre
1 mL = 1 cm3 (= 1cc)
Other metric units of measuring volume are given in the table below:
large volume → → → → → → → → → → small volume
name teralitre gigalitre megalitre kilolitre hectolitre decalitre decilitre centilitre millilitre microlitre nanolitre picolitre femtolitre attolitre
symbol TL GL ML kL hL daL dL cL mL µL nL pL fL aL
volume (L) 1012 L 109 L 106 L 103 L 102 L 101 L 10-1 L 10-2 L 10-3 L 10-6 L 10-9 L 10-12 L
Answer:
The rate the radius of the balloon shrinking at the moment the radius is 5 centimeters is 0.1783 cm/s
Explanation:
Here we have
dV/dt = 56 cm³/s
When the radius is 5 cm we have
Therefore,
From which,
The rate the radius of the balloon shrinking at the moment the radius is 5 centimeters = 0.1783 cm/s.