1000 cm3 of air at 20 temperature and 101.35 kpa is heated at constant pressure until its volume doubles.
a, use the ideal gas equation to calculate the final temperature of the gas.
b, calculate work done by the gas as it expands.
a) To calculate the final temperature of the gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure = 101.35 kPa
V = volume = 1000 cm3
n = number of moles of gas (constant for a given amount of gas)
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature in Kelvin (K)
First, we need to convert the volume from cm3 to liters:
V = 1000 cm3 * (1 L / 1000 cm3) = 1 L
We know that the volume doubles, so the final volume, Vf = 2 * V = 2 L.
Next, let's write the equation for the initial state of the gas:
(P1)(V1) = (n)(R)(T1)
And for the final state of the gas:
(P2)(V2) = (n)(R)(T2)
Since the pressure is constant, P1 = P2 = P = 101.35 kPa.
We can rearrange the equations to solve for the final temperature, T2:
(V1)(T1) = (V2)(T2)
(1 L)(T1) = (2 L)(T2)
T2 = (1 L * T1) / 2 L
Since T1 is given as 20 degrees Celsius, we need to convert it to Kelvin:
T1 = 20 + 273.15 = 293.15 K
Plugging in the values, we get:
T2 = (1 L * 293.15 K) / 2 L
T2 = 146.57 K
Therefore, the final temperature of the gas is approximately 146.57 K.
b) To calculate the work done by the gas as it expands, we can use the formula:
Work = P * ΔV
Where:
P = pressure = 101.35 kPa
ΔV = change in volume = Vf - Vi
From part a, we know that Vi = 1 L and Vf = 2 L, so:
ΔV = 2 L - 1 L = 1 L
Plugging in the values, we get:
Work = 101.35 kPa * 1 L
Work = 101.35 Joules (J)
Therefore, the work done by the gas as it expands is approximately 101.35 J.
To solve this problem, we can use the ideal gas equation, which states:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature
Given:
Initial volume, V1 = 1000 cm^3
Initial temperature, T1 = 20 degrees Celsius = 20 + 273.15 = 293.15 K
Initial pressure, P1 = 101.35 kPa
a) To calculate the final temperature of the gas after doubling its volume, we need to rearrange the ideal gas equation and solve for T2.
From the ideal gas equation, we can rewrite it as:
(P1V1) / T1 = (P2V2) / T2
Since the pressure is constant, we can cancel it out:
(V1) / T1 = (V2) / T2
Substituting the known values:
(1000 cm^3) / 293.15 K = (2 * 1000 cm^3) / T2
Now, we can solve for T2:
T2 = (2 * 1000 cm^3 * 293.15 K) / 1000 cm^3
T2 = 586.3 K
Therefore, the final temperature of the gas is 586.3 K.
b) To calculate the work done by the gas as it expands, we can use the formula:
Work (W) = P * ΔV
Where:
P is the constant pressure
ΔV is the change in volume
In this case, the gas doubles its volume, so:
ΔV = 2 * V1 = 2 * 1000 cm^3 = 2000 cm^3
Substituting the known values:
W = 101.35 kPa * 2000 cm^3
To convert cm^3 to liters, we divide by 1000:
W = 101.35 kPa * (2000 cm^3 / 1000)
W = 202.7 kPa * cm^3
The unit of work in the SI system is joules (J), so we need to convert kPa * cm^3 to joules.
1 kPa * cm^3 = 1 J
Therefore, the work done by the gas as it expands is 202.7 J.
energy=PV
work done= change in PV=101.3kps*(1e-6m^3)=101.4e-9 N/m^2 * m^3=
101.3e-9 joules
V1/V2=T1/T2
V1=1000
T1=273+20 (Kelivins)
V2=2000
T2=2(273+20) Kelvins