Work done isothermal, adiabatic ideal gas

[Samuelriesterer]

Problem statement, work done, and relevant equations:

One mole of ideal gas is initially at 1 atm and has a volume of 5L.

a) Calculate the work done on the gas during an isothermal, reversible compression to a volume of 2L.

$W_isothermal = - \int_{v_i}^{v_f} p dv = - \int_{v_I}^{v_f} \frac{nRT}{V} dV = -nRT \int_{v_I}^{v_f} \frac{1}{V} dV = -nRT ln \frac{V_I}{V_f} = -(1 mol)(8.314 \frac{J}{molK})T ln \frac{.005 m^3}{.002 m^3} = 7.618T$

b) Calculate the work done during a reversible adiabatic compression to 2L.

$P_i = \frac{nRT_I}{V} → T_i = \frac{V_I P_i}{nR} = \frac{(.005 m^3)(101300 Pa)}{(1 mol)(8.314 \frac{J}{molK})} = 60.92 K$

I think I'm a little stuck here because the problem statement does not give me whether its a monatomic or diatomic gas in order to get the $\frac{C_p}{C_v}$ value. Is there something I'm not seeing?

c) What are the final pressures for (a) and (b)?

 

[Andrew Mason]

a) Calculate the work done on the gas during an isothermal, reversible compression to a volume of 2L.

$W_isothermal = - \int_{v_i}^{v_f} p dv = - \int_{v_I}^{v_f} \frac{nRT}{V} dV = -nRT \int_{v_I}^{v_f} \frac{1}{V} dV = -nRT ln \frac{V_I}{V_f} = -(1 mol)(8.314 \frac{J}{molK})T ln \frac{.005 m^3}{.002 m^3} = 7.618T$

b) Calculate the work done during a reversible adiabatic compression to 2L.

$P_i = \frac{nRT_I}{V} → T_i = \frac{V_I P_i}{nR} = \frac{(.005 m^3)(101300 Pa)}{(1 mol)(8.314 \frac{J}{molK})} = 60.92 K$

I think I'm a little stuck here because the problem statement does not give me whether its a monatomic or diatomic gas in order to get the $\frac{C_p}{C_v}$ value. Is there something I'm not seeing?

The first part is correct and your determination of $T_i$ is correct, assuming you have done the arithmetic properly. You do need to know the value for $\gamma = \frac{C_p}{C_v}$ in order to calculate the adiabatic work and the final pressure and temperature (applying the adiabatic condition). I would suggest that you assume a true ideal gas (monatomic).

AM