A sample of 8.02×10^−1 moles of nitrogen gas (γ =1 .40) occupies a volume of 2.00×10^−2 m^3 at a pressure of 1.00×10^5 Pa and temperature of 300K. The sample is isothermally compressed to exactly half its original volume. Nitrogen behaves as an ideal gas under these conditions.
(i)Find the change in entropy of the gas. (ii) What is the change in internal energy of the gas?
(iii) What is the amount of heat transferred from the gas to its environment?
(iv) Calculate the amount of work done in compressing the gas
Anyone got any ideas? I think I can find the work done but im not sure about the other questions.
i) the change in entropy is zero as its isothermal.
ii) Also zero.
iii) use P deltaV=q.
To find the change in entropy of the gas, we can use the equation:
ΔS = nR ln(V₂/V₁)
where:
ΔS is the change in entropy
n is the number of moles of gas
R is the gas constant (8.314 J/(mol·K))
V₁ is the initial volume
V₂ is the final volume
First, let's calculate the initial volume (V₁) using the given information:
V₁ = 2.00×10^(-2) m^3
Then, since the sample is isothermally compressed to exactly half its original volume, the final volume (V₂) is:
V₂ = 0.5 * V₁ = 0.5 * 2.00×10^(-2) = 1.00×10^(-2) m^3
Next, we need to calculate the number of moles of nitrogen gas (n):
n = 8.02×10^(-1) moles
Now, we can substitute these values into the equation to find the change in entropy (ΔS):
ΔS = (8.02×10^(-1) moles) * (8.314 J/(mol·K)) * ln(1.00×10^(-2) m^3 / 2.00×10^(-2) m^3)
Simplify the equation and calculate the value of ΔS.
To find the change in internal energy of the gas, we can use the equation:
ΔU = Q - W
where:
ΔU is the change in internal energy
Q is the heat transferred to the gas
W is the work done on the gas
To calculate ΔU, we need to find Q and W.
(iii) To find the amount of heat transferred (Q), we can use the equation:
Q = nCΔT
where:
C is the molar specific heat capacity of the gas
ΔT is the change in temperature
Given that the sample is isothermal (at a constant temperature of 300K), there is no change in temperature (ΔT = 0). Therefore, Q = 0 J.
(iv) To calculate the work done (W), we can use the equation:
W = -PΔV
where:
P is the pressure of the gas
ΔV is the change in volume
Given that the gas is compressed, ΔV is negative:
ΔV = V₂ - V₁ = (1.00×10^(-2) m^3) - (2.00×10^(-2) m^3) = -1.00×10^(-2) m^3
Substitute the values into the equation to find the work done (W).