A sample of monatomic gas (£^ = 5/3) is expanded adiabatically from an initial temperature, volume, and pressure of 230. ¢XC, 6.00 litres, and 5.00 kPa, respectively, to a final volume of 7.00 litres.

(a) What is the final pressure of the gas?
(b) How much work is done by the gas during the adiabatic expansion?

dude i have no idea!

(a) During the expansion, the product

P*V^(5/3) remains constant. You know the ratio V2/V1. Use the adiabatic relationship to calculate P2/P1.

P2/P1 = (V1/V2)^(5/3)
= (6/7)^(5/3) = 0.773

(2) The work done is the integral of P dV from V1 to V2. Express P as a function of V and integrate

P(V) dV = [P1*V1^(5/3)]*/V^5/3)]dV

To answer these questions, we can use the adiabatic expansion equation for an ideal gas:

\(P_1 V_1^{\gamma} = P_2 V_2^{\gamma}\)

Where:
\(P_1\) = Initial pressure
\(V_1\) = Initial volume
\(P_2\) = Final pressure
\(V_2\) = Final volume
\(\gamma\) = Heat capacity ratio for a monatomic gas (\(\gamma = \frac{5}{3}\))

(a) To find the final pressure of the gas (P2), we can rearrange the equation as:

\(P_2 = P_1 \left(\frac{V_1}{V_2}\right)^{\gamma}\)

Substituting the given values:
\(P_1 = 5.00 \, \text{kPa}\)
\(V_1 = 6.00 \, \text{litres}\)
\(V_2 = 7.00 \, \text{litres}\)
\(\gamma = \frac{5}{3}\)

\(P_2 = 5.00 \, \text{kPa} \left(\frac{6.00 \, \text{litres}}{7.00 \, \text{litres}}\right)^{\frac{5}{3}}\)

Calculating this expression gives us the final pressure of the gas.

(b) To find the work done by the gas during the adiabatic expansion, we can use the following equation:

\(W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1}\)

Substituting the given values:
\(P_1 = 5.00 \, \text{kPa}\)
\(V_1 = 6.00 \, \text{litres}\)
\(P_2\) (from part a)
\(V_2 = 7.00 \, \text{litres}\)
\(\gamma = \frac{5}{3}\)

Calculating this expression gives us the work done by the gas during the adiabatic expansion.