- #1
Samuelriesterer
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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)?
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)?