An ideal monatomic gas expands isothermally from 0.590 m3 to 1.25 m3 at a constant temperature of 670 K. If the initial pressure is 8.00 104 Pa.
(a) Find the work done on the gas.
Answer is in J
(b) Find the thermal energy transfer Q.
Answer is in J
(a) Well, when it comes to work, this monatomic gas is quite the professional. To find the work done on the gas during this isothermal expansion, you can use the formula:
W = nRT ln(Vf/Vi)
Where:
W is the work done
n is the number of moles of gas
R is the ideal gas constant
T is the temperature in Kelvin
Vf is the final volume
Vi is the initial volume
But since we're dealing with a monatomic gas, we can simplify things a bit. For a monatomic gas, the number of moles is given by n = pV / RT, where p is the pressure. So, we can rewrite the formula as:
W = pDeltaV
Where DeltaV is the change in volume, which is Vf - Vi.
Now, to the calculations: Delta V is 1.25 m^3 - 0.590 m^3 = 0.660 m^3. Plug in the values:
W = (8.00 * 10^4 Pa) * (0.660 m^3)
W = 5.28 * 10^4 J
So, the work done on the gas is approximately 5.28 * 10^4 J.
(b) Ah, thermal energy transfer Q, we meet again! In an isothermal process, the thermal energy transferred is equal to the work done on the gas. So, Q = W.
Therefore, the thermal energy transfer Q is also approximately 5.28 * 10^4 J.
Well, that's the end of this gas's isothermal expansion journey. It's a gas that really knows how to work it out!
To find the work done on the gas during the isothermal expansion, we can use the formula:
Work = -nRT * ln(V2/V1)
where:
n is the number of moles of the gas
R is the ideal gas constant (8.314 J/(mol*K))
T is the temperature in Kelvin
V1 and V2 are the initial and final volumes of the gas, respectively
First, let's calculate the number of moles of the gas using the ideal gas law:
PV = nRT
Rearranging the equation:
n = PV / RT
Given:
P = 8.00 * 10^4 Pa
V1 = 0.590 m^3
V2 = 1.25 m^3
T = 670 K
Substituting these values into the equation:
n = (8.00 * 10^4 Pa * 0.590 m^3) / (8.314 J/(mol*K) * 670 K)
n ≈ 65.57 mol
Now, let's calculate the work done:
Work = -(65.57 mol * 8.314 J/(mol*K) * 670 K) * ln(1.25 m^3 / 0.590 m^3)
Work ≈ -2.22 * 10^5 J
(a) The work done on the gas during the isothermal expansion is approximately -2.22 * 10^5 J.
Now let's move on to part (b), finding the thermal energy transfer Q.
For an isothermal process, Q (thermal energy transfer) is equal to the work done:
Q = -2.22 * 10^5 J
(b) The thermal energy transfer Q during the isothermal expansion is approximately -2.22 * 10^5 J.
To find the work done on the gas during the isothermal expansion, we can use the formula:
Work (W) = nRT * ln(V2/V1)
where:
- n is the number of moles of the gas
- R is the ideal gas constant (8.31 J/(mol·K))
- T is the temperature of the gas in Kelvin
- V2 and V1 are the final and initial volumes of the gas respectively
First, we need to find the number of moles of the gas. We can use the ideal gas law equation:
PV = nRT
Rearranging the equation, we get:
n = PV / RT
Given:
- P = 8.00 x 10^4 Pa
- V1 = 0.590 m^3
- V2 = 1.25 m^3
- T = 670 K
Plug in the values to calculate the number of moles (n):
n = (8.00 x 10^4 Pa * 0.590 m^3) / (8.31 J/(mol·K) * 670 K)
Now that we have the value of n, we can calculate the work done:
W = nRT * ln(V2/V1)
W = n * (8.31 J/(mol·K)) * 670 K * ln(1.25 m^3 / 0.590 m^3)
Calculate the value of W to find the answer in Joules.
To find the thermal energy transfer Q, which represents the heat exchanged during the process, we can use the equation:
Q = W
Since the expansion is isothermal, the change in internal energy of the gas is zero, resulting in no change in thermal energy. Thus, the thermal energy transfer (Q) is equal to the work done (W).
Therefore, the answer for both parts (a) and (b) will be the same and can be obtained from the calculation above.
(a) Work out = Wout
= Integral of P dV = nRT/V *dV
= nRT*ln(V2/V1)
The initial pressure and ideal gas law will tell you what n is
(b) Qin = Wout , since the internal enery (and T) do not change