Question

A sample of an ideal gas at 15.0 atm and 10.0 l is allowed to expand against a constant external pressure of 2.00 atm at a constant temperature. calculate the work in units of kj for the gas expansion.

Answer 1

According to Boyle's law the volume of a gas is inversely proportional to its pressure. Boyle's law follows the equation:  

P1*V1 = P2*V2 P1= 15atm V1= 10L P2= 2atm V2= ?  

(15atm)*(10L)= (2atm) *(V2)  

V2= 75L  

change in volume = final volume - initial volume =  

75L - 10L = 65L  

The work done by an expanding gas can be calculated by the equation:  

w = -(Pressure)*(change in volume)  

The pressure is the amount of pressure under which the gas expands which is 2atm (If the pressure was 15atm, the gas would not expand). So:  

w = -(2atm)*(65L) = -130 L*atm  

Now you have to convert from L*atm to J. Here's how you do it. The conversion factor can be obtained from R values.  

.08206 (L*atm)/(K*mol) and 8.314 J/(K*mol)  

We want J/(L*atm) so divide the R values the denominators cancel:  

(8.314J/(K*mol)) / ((.08206L*atm)/(K*mol)) = 101.3J/(L*atm)  

See how the K*mol cancels.  

w= -130L*atm *(101.3J/(L*atm)) *(kJ/1000J) =  

w= -13.2kJ

Answer 2

The work for the gas expansion from a sample of an ideal gas at 15.0 atm and 10.0 L that is allowed to expand against a constant external pressure of 2.00 atm at a constant temperature is = 13.2 kJ

The given parameters include:

• Initial pressure = 15.0 atm
• Initial volume = 10.0 L
• Final external pressure = 2.00 atm

From above, we can determine the Final volume by using the equation for  Boyle's Law.

Using the equation:

P₁V₁ = P₂V₂

(15.0 atm × 10.0 L) = (2.00 atm × V₂)

Making V₂ the subject;

$\mathbf{V_2 = \dfrac{(15.0 \ atm \times 10.0 \ L)}{2.00 \ atm}}$

V₂ = 75.0 L

The change in the volume ΔV = V₂ - V₁

• ΔV = 75.0 L - 10.0 L
• ΔV = 65.0 L

However, since the temperature is constant and the gas performs work on the surroundings.

The Work of the gas expansion is:

• W = -PΔV

where;

• P = external pressure = 2.00 atm

∴

W = -(2.00 atm × 65.0 L)

W = -130 L. atm

$\mathbf{(W = -130 L. atm \times (\dfrac{101.3 \ J}{1 \ L.atm})}$

W = 13169 J

W = 13.2 kJ

Therefore, we can conclude that the work for the gas expansion is 13.2 kJ

Learn more about Boyle's Law here:

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