Question

A piston/cylinder is placed in a constant-temperature bath and the gas is held at a pressure of 11,01 bar and initial volume of 0.02 m^3, by an external force. If the external force is gradually reduced so that the gas expands isothermally and reversibly by doing 25,892] of work on the surrounding air, what is its final volume in m^3, to 4 decimal places? Assume that the total gas volume is related to its pressure by PV =k, where k is a constant.

Answer 1

In this scenario, we are working with an isothermal (constant temperature) and reversible process. The equation for the work done during such a process in ideal gases is given by:

W = P_initial * V_initial * ln(V_final/V_initial)

Here, the work W is given as 25,892 J, the initial pressure P_initial is 11.01 bar (which is 1101 kPa or 1101000 Pa to convert it into SI units), and the initial volume V_initial is 0.02 m^3. We want to find the final volume V_final. We can rearrange the equation to solve for V_final:

V_final = V_initial * exp(W / (P_initial * V_initial))

Substituting in the given values:

V_final = 0.02 m^3 * exp(25892 J / (1101000 Pa * 0.02 m^3))

To calculate the exponential function, you can use a calculator. We can compute:

exp(25892 / (1101000 * 0.02)) = exp(1.177)

V_final = 0.02 m^3 * exp(1.177)

After evaluating the exponential, we get:

V_final = 0.02 m^3 * 3.247 = 0.06494 m^3

So, the final volume is approximately 0.0649 m^3 to four decimal places.