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Part 3/3 find the thermal energy transferred. Answer in units of KJ

Part 33 Find The Thermal Energy Transferred Answer In Units Of KJ class=

Sagot :

0.0We are asked to determine the volume of a gas given the moles, the temperature, and the pressure. To do that, we use the following formula:

[tex]PV=nRT[/tex]

Where:

[tex]\begin{gathered} P=pressure \\ V=\text{ volume} \\ n=\text{ number of moles} \\ R=\text{ universal gas constant} \\ T=\text{ temperature} \end{gathered}[/tex]

Now, we substitute the values for the final state. Since the process is isothermally this means that the final temperature is the same as the initial temperature. we get:

[tex](1.8atm)V_f=(2mol)(8.31415\frac{J}{Kmol})(243K)[/tex]

We need to convert the pressure from atmospheres to Pascals. To do that we use the following conversion factor:

[tex]1atm=101325Pa[/tex]

Multiplying by the conversion factor we get:

[tex]1.8atm\times\frac{101325Pa}{1atm}=182385Pa[/tex]

Now, we substitute the value in the formula:

[tex](182385Pa)V_f=(2mol)(8.3145\times\frac{J}{Kmol})(243K)[/tex]

Solving the operations:

[tex](182385Pa)V_f=4040.847J[/tex]

Now, we divide both sides by 182385Pa:

[tex]V_f=\frac{4040.847J}{182385Pa}[/tex]

Solving the operations:

[tex]V_f=0.022m^3[/tex]

Therefore, the final volume is 0.022 cubic meters.

Part B. We are asked to determine the work done. To do that we will use the formula for isothermic work:

[tex]W=nRT\ln(\frac{P_0}{P_f})[/tex]

Where:

[tex]P_0,P_f=\text{ initial and final pressure}[/tex]

Now, we plug in the values:

[tex]W=(2mol)(8.31451\frac{J}{Kmol})(243K)\ln(\frac{0.21atm}{1.8atm})[/tex]

Now, we solve the operations:

[tex]W=-8681.51J[/tex]

Therefore, the work done is -8681.51 Joule. To convert to kilojoules we divide by 1000:

[tex]W=-8681.51J\times\frac{1kJ}{1000J}=-8.68kJ[/tex]

Part 3. In an isothermic process the change in internal energy is zero, therefore, according to the first law of thermodynamics we have:

[tex]Q-W=0[/tex]

Therefore:

[tex]Q=W[/tex]

Therefore, the amount of heat is equal to the amount of work. Therefore, the thermal energy transferred is:

[tex]Q=-8.68kJ[/tex]

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