A monatomic ideal gas expands from 1.00m^3 to 2.50m^3 st a constant pressure of 2.00x10^5 Pa. Find the change in the internal energy of the gas.
Internal energy is U=3•n•R•T/2 => chage in internal energy is ΔU=3•n•R•ΔT/2.
From the ideal gas law PV=nRT =>
n•R•ΔT=nR(T2-T1)=p2•V2 =p1•V1=p(V2-V1) =>
ΔU=3•n•R•ΔT/2=3•p(V2-V1)/2=3•2•10⁵•(2.5-1)/2=4.5•10⁵ J
To find the change in the internal energy of the gas, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system:
ΔU = Q - W
In this case, the gas expands at a constant pressure, so the work done by the system can be expressed as:
W = PΔV
where P is the pressure and ΔV is the change in volume.
Given:
Initial volume, V1 = 1.00 m^3
Final volume, V2 = 2.50 m^3
Pressure, P = 2.00 × 10^5 Pa
The change in volume is calculated as:
ΔV = V2 - V1
= 2.50 m^3 - 1.00 m^3
= 1.50 m^3
Now we can substitute these values into the equation for work:
W = PΔV
W = (2.00 × 10^5 Pa)(1.50 m^3)
W = 3.00 × 10^5 J
Since the process is isobaric (constant pressure), the change in internal energy can be calculated as:
ΔU = Q - W
Since no heat is added or taken from the system (Q = 0), the change in internal energy is simply equal to the work done by the system:
ΔU = W
ΔU = 3.00 × 10^5 J
Therefore, the change in the internal energy of the gas is 3.00 × 10^5 J.
To find the change in internal energy of the gas, we can use the First Law of Thermodynamics, which states that the change in internal energy (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W).
In this case, the gas is expanding at a constant pressure, so the work done by the gas is given by the formula W = PΔV, where P is the pressure and ΔV is the change in volume.
Given:
Initial volume (V1) = 1.00 m^3
Final volume (V2) = 2.50 m^3
Pressure (P) = 2.00 x 10^5 Pa
First, calculate the change in volume:
ΔV = V2 - V1
ΔV = 2.50 m^3 - 1.00 m^3
ΔV = 1.50 m^3
Next, calculate the work done by the gas:
W = PΔV
W = (2.00 x 10^5 Pa)(1.50 m^3)
W = 3.00 x 10^5 J
Since the pressure is constant, the heat added to the system (Q) is equal to the work done by the gas. So, Q = W.
Finally, calculate the change in internal energy:
ΔU = Q - W
ΔU = (3.00 x 10^5 J) - (3.00 x 10^5 J)
ΔU = 0 J
Therefore, the change in internal energy of the gas is 0 J.