Question

Two gases have same initial pressure, volume and temperature. They expand to the same final volume,one adiabatically and the other isothermally
(1) The final pressure is greater for isothermal process
(2) The final temperature is greater for the isothermal process
(3) The work done by the gas is greater for the isothermal process
(4) All of these

Answer 1

Answer on Question #41254 – Physics - Molecular Physics

Question.

Two gases have same initial pressure, volume and temperature. They expand to the same final volume, one adiabatically and the other isothermally

(1) The final pressure is greater for isothermal process

(2) The final temperature is greater for the isothermal process

(3) The work done by the gas is greater for the isothermal process

(4) All of these

$P_{1}$ is an initial pressure

$V_{1}$ is an initial volume

$T_{1}$ is an initial temperature

$P_{end}$ is the final pressure

$T_{end}$ is an initial temperature

Solution.

By definition work done is:

$$\delta A = P d V$$ $$A = \int \delta A = \int P d V$$

Adiabatic process:

$$T V ^ {\gamma - 1} = c o n s t$$ $$T _ {1} V _ {1} ^ {\gamma - 1} = T V ^ {\gamma - 1} \rightarrow T _ {e n d} = T _ {1} \left(\frac {V _ {1}}{V}\right) ^ {\gamma - 1}$$ $$P V ^ {\gamma} = c o n s t$$ $$P _ {1} V _ {1} ^ {\gamma} = P V ^ {\gamma} \rightarrow P _ {e n d} = P _ {1} \left(\frac {V _ {1}}{V}\right) ^ {\gamma}$$ $$A = \int \delta A = \int P d V = \int_ {V _ {1}} ^ {V _ {2}} P _ {1} V _ {1} ^ {\gamma} \frac {d V}{V ^ {\gamma}} = \gamma P _ {1} V _ {1} ^ {\gamma} \left(\frac {1}{V _ {1} ^ {\gamma - 1}} - \frac {1}{V _ {2} ^ {\gamma - 1}}\right) = \gamma P _ {1} V _ {1} \left(1 - \left(\frac {V _ {1}}{V _ {2}}\right) ^ {\gamma - 1}\right)$$

Isothermal process:

$$T _ {e n d} = T _ {1}$$ $$P V = R T _ {1}$$ $$P = \frac {R T _ {1}}{V} \rightarrow P _ {e n d} = \frac {P _ {1} V _ {1}}{V}$$ $$A = \int \delta A = \int P d V = \int_ {V _ {1}} ^ {V _ {2}} R T _ {1} \frac {d V}{V} = R T _ {1} \ln \left(\frac {V _ {2}}{V _ {1}}\right) = P _ {1} V _ {1} \ln \left(\frac {V _ {2}}{V _ {1}}\right)$$

So,

$$P _ {e n d} ^ {i s o t h e r m} = \frac {P _ {1} V _ {1}}{V} > P _ {e n d} ^ {a d i a b a t i c} = P _ {1} \left(\frac {V _ {1}}{V}\right) ^ {\gamma} \leftrightarrow P _ {e n d} ^ {i s o t h e r m} > P _ {e n d} ^ {a d i a b a t i c}$$ $$T _ {e n d} ^ {i s o t h e r m} = T _ {1} > T _ {e n d} ^ {a d i a b a t i c} = T _ {1} \left(\frac {V _ {1}}{V}\right) ^ {\gamma - 1} \leftrightarrow T _ {e n d} ^ {i s o t h e r m} > T _ {e n d} ^ {a d i a b a t i c}$$

$A = \int P d V$, therefore the work done by the gas is greater for process with greater final pressure and temperature $\rightarrow$ The work done by the gas is greater for the isothermal process.

You can also see it from:

$$A ^ {i s o t h e r m} = P _ {1} V _ {1} \ln \left(\frac {V _ {2}}{V _ {1}}\right) > A ^ {a d i a b a t i c} = \gamma P _ {1} V _ {1} \left(1 - \left(\frac {V _ {1}}{V _ {2}}\right) ^ {\gamma - 1}\right) \leftrightarrow A ^ {i s o t h e r m} > A ^ {a d i a b a t i c}$$

Thus, the final pressure, temperature and work done by the gas are greater for isothermal process than for adiabatic process.

Answer.

(4) All of these

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