Question

A cylinder with a movable piston contains a sample of ideal gas. A temperature probe and a pressure probe are inserted into the cylinder, as shown above. The sample of gas is taken through a three-step cycle, ABCA In process AB, the volume is decreased to 1/4 its original value while constant pressure is maintained

Answer 1

Answer:

i. The values for the graph are;

Volume Pressure

1                  1

1/4               1

1                   1/4

1                   1

The graph is attached

ii. Energy is added to the system.

Explanation:

i) Here we note that Charles law states that the volume of a given mass of gas is directly proportional to its temperature at constant pressure.

Therefore;

T₁/V₁ = T₂/V₂

Where V₂ = V₁/4, we have;

T₁/V₁ = T₂/(V₁/4) = 4×T₂/V₁

T₁ =  4×T₂

Therefore, T₂ = T₁ /4

In process BC, whereby the gas expands isothermally to original volume, we have; according to Boyle's law

P₁·V₁  = P₂·V₂

V₁ = 4·V₂

∴ P₁·4·V₂  = P₂·V₂

4·P₁  = P₂

Process CA, gas returns to initial state with T₂ = T₁ /4

Therefore, we have from Pₐ·Vₐ/Tₐ = P₁·V₁/T₁

Where the volume remain constant, we have Gay Lussac's Law which states that the pressure of a given mass of gas is directly proportional to its temperature, therefore

P₁/T₁ = P₂/T₂

Hence; whereby the temperature changes from T₂ to T₁ and T₂ = T₁ /4, we have P₁ = 4·P₂

Therefore, the pressure increases by a factor of 4 × 4 = 16.

Please find the graph attached

ii) Given that the gas expands to its original volume V₁, such that V₁ = 4·V₂, also work done, W is given as follows;

$W = nRTln\frac{v_f}{v_i} = nRTln\frac{v_1}{v_2} \ \because v_f =v_1 \ and \ v_i = v_2 \ also \ v_1 = 4\cdot v_2$

Therefore;

$W = nRTln\frac{4\cdot v_2}{v_2} \ \therefore W = nRTln(4)$

Hence, as n, R, and T are all positive values, W is positive, energy is added to the system.