five mole of an ideal gas are kept at a constant temperature of 53 degree celcius while the pressure of the gas is increased from 1 atm to 3 atm calculate the work done by the gas
Your ans is wrong
Well, well, well, looks like we have a gas party going on! Let's calculate how much work this jovial gas has done.
To find the work done by the gas, we can use the formula:
Work = -P * ΔV
Where P is the pressure and ΔV is the change in volume. Since the temperature is constant, we can use Boyle's Law to find the change in volume.
Boyle's Law states that for an ideal gas at a constant temperature, the product of pressure and volume is constant. So we can write:
P1 * V1 = P2 * V2
We are given that P1 = 1 atm, V1 is the initial volume, P2 = 3 atm, and we need to find V2.
Now, since we have moles of the gas, we can use the ideal gas law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/(mol·K)) and T is the temperature in Kelvin.
Now, let's get down to business:
First, convert the temperature from Celsius to Kelvin: 53 °C + 273.15 = 326.15 K
We are given that n = 5 moles, R = 0.0821 L·atm/(mol·K), P1 = 1 atm, and T = 326.15 K. Plug these values into the ideal gas law to find V1:
(1 atm) * V1 = (5 moles) * (0.0821 L·atm/(mol·K)) * (326.15 K)
Solve for V1 and you'll find V1 = (5 moles) * (0.0821 L·atm/(mol·K)) * (326.15 K) / (1 atm).
Now, we can use Boyle's Law to find V2:
(1 atm) * V1 = (3 atm) * V2
Plug in the values and solve for V2. You've got this!
Finally, use the formula for work:
Work = -P * ΔV
Plug in the values of P2 and V2, and don't forget the negative sign!
Voila! You've calculated the work done by the gas! I hope this gas had a good workout.
To calculate the work done by the gas, we can use the formula:
Work = PΔV
Where:
P = pressure change (in this case, it is the difference between the final pressure and the initial pressure)
ΔV = change in volume
Given:
Initial pressure (P1) = 1 atm
Final pressure (P2) = 3 atm
Let's calculate the change in pressure (ΔP) and the change in volume (ΔV).
ΔP = P2 - P1
= 3 atm - 1 atm
= 2 atm
We know that 1 atm = 101.325 joules/liter, so we can convert the pressure change from atm to joules/liter.
ΔP = 2 atm x 101.325 joules/liter/atm
= 202.65 joules/liter
Now, to calculate the change in volume, we can use the ideal gas law:
PV = nRT
Where:
P = pressure (in this case, it is the initial pressure)
V = volume
n = number of moles
R = ideal gas constant (0.0821 atm⋅L/mol⋅K)
T = temperature (in Kelvin)
Let's calculate the change in volume (ΔV):
ΔV = V2 - V1
= (nRT2)/P2 - (nRT1)/P1
= nR(P2T2 - P1T1)/(P1P2)
= (5 mol)(0.0821 atm⋅L/mol⋅K)((3 atm)(326 K) - (1 atm)(326 K))/((1 atm)(3 atm))
≈ 2.194 L
Now we can calculate the work done by the gas:
Work = PΔV
= (202.65 joules/liter)(2.194 L)
≈ 445.47 joules
Therefore, the work done by the gas is approximately 445.47 joules.
To calculate the work done by the gas, we can use the formula:
Work = -PΔV
Where:
Work = Work done by the gas
P = Change in pressure
ΔV = Change in volume
In this case, we know that the pressure changes from 1 atm to 3 atm, which is an increase of 2 atm. However, we don't have the value of ΔV yet. To find it, we can use the ideal gas law:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal gas constant
T = Temperature in Kelvin
Let's convert the given temperature from Celsius to Kelvin:
T(K) = 53°C + 273.15 = 326.15 K
Now we can rearrange the ideal gas law to find ΔV:
ΔV = (nRΔT) / P
Where:
ΔV = Change in volume
n = Number of moles
R = Ideal gas constant
ΔT = Change in temperature
P = Pressure
Substituting the given values:
ΔV = (5 mol * 0.0821 L*atm/mol*K * 326.15 K) / 1 atm
Calculating this equation gives us the ΔV value. Once we have the ΔV value, we can substitute it into the formula for work and multiply it by -1 to get the final value of work done.
At constant T, Volume V is inversely proportional to P. Therefore
PV = P1*V1
Compute V1 first of all
V1 = nRT/P1 = 134 liters = 0.134 m^3
P1 = 1 atm = 1.013*10^5 N/m^2
Work done on gas =
-Integral of P dV = -P1*V1 Integral of dV/V
= P1*V1 ln (V1/V2) = P1*V1*ln(P2/P1)
= P1*V1 ln3 = 1.49*10^4 Joules