Question

$n$moles of an ideal gas at temperature $T_1$and volume localid="1648032029846" $V_1$expand isothermally until the volume has doubled. In terms of $n,T_1$and $V_1$, what are (a) the final temperature, (b) the work done on the gas, and (c) the heat energy transferred to the gas?

Short answer

a) The final temperature is equal to the initial temperature $T_1.$

b) The work done on the gas is $W=-nR{T}_1\ln \left(2\right)$

c) The heat energy transferred to the gas is$Q=nR{T}_1\ln \left(2\right)$

Step-by-step solution

Given Information (Part a)

Number of moles of an ideal gas$=n$

Temperature$=T_1$

Volume$=V_1$

Explanation (Part a)

Since the process is isothermic, the final temperature will be equal to the initial one,$T_1.$

Final Answer (Part a)

Hence, the initial temperature is equal to the final temperature.

Given Information (Part b)

Number of moles of an ideal gas $=n$

Temperature $=T_1$

Volume$=V_1$

Explanation (Part b)

(b) The work in an isothermic process is given by

$W=-nRT\ln \left(\frac{V_2}{V_1}\right)$

Since the volume was doubled, in our case we simply have

$W=-nR{T}_1\ln \left(2\right)$

Final Answer (Part b)

Hence, the work done on the gas is$W=-nR{T}_1\ln \left(2\right).$

Given Information (Part c)

Number of moles of an ideal gas $=n$

Temperature $=T_1$

Volume $=V_1$

Explanation (Part c)

(c) Since the process is isothermal, the internal energy of the gas didn't change.

This means that, from the first law, the entire heat added to it was converted into work. Therefore, the heat added to the gas was

$Q=nR{T}_1\ln \left(2\right)$

Final Answer (Part c)

Hence, the heat energy transferred to the gas is$Q=nR{T}_1\ln \left(2\right).$