Question

$5.0\mathrm{g}$of nitrogen gas at $20°\mathrm{C}$and an initial pressure of $3.0\mathrm{atm}$undergo an isobaric expansion until the volume has tripled.

a. What are the gas volume and temperature after the expansion?

b. How much heat energy is transferred to the gas to cause this expansion?

The gas pressure is then decreased at constant volume until the original temperature is reached.

c. What is the gas pressure after the decrease?

d. What amount of heat energy is transferred from the gas as its pressure decreases?

e. Show the total process on a $pV$diagram. Provide an appropriate scale on both axes.

Short answer

a) The gas volume and temperature after the expansion is$4300c{m}^3$,${606}^{\circ }\mathrm{C}$

b) The amount of heat energy is transferred to the gas to cause this expansion is $3050J$

c) The gas pressure after the decrease is $p_3=1atm$

d) The amount of heat energy is transferred from the gas as its pressure decreases is $-2180J.$

e) The total process on a $pV$diagram is as follows:

Step-by-step solution

Given Information (Part a)

Amount of nitrogen gas$=5.0g$

Temperature $={20}^{\circ }C$

Initial pressure$=3.0atm$

Explanation (Part a)

(a) Given that the volume after the expansion is three times the initial one, we just need to find the latter. From the ideal gas law, we have

$p_1{V}_1=nR{T}_1\Rightarrow V_1=\frac{nR{T}_1}{p_1}$

Knowing that the number of moleslocalid="1648620030592" $n$is given as $n=\frac{m}{M}$, where $m$is the mass and $M$is the molar mass, we have

localid="1648620200691" $V_1=\frac{mR{T}_1}{p_{1}M}=\left(\frac{0.005kg\times 8.314kg\cdot m^2\cdot s^{-2}\cdot K^{-1}\cdot mo{l}^{-1}\times 293K}{3atm\times 1.01\times {10}^5·0.028kg/mol}\right)$

Multiply the values,

$=1.44·{10}^{-3}{\mathrm{m}}^3$

$=1.44L$

Therefore, the volume after the expansion is

$V_2=3·V_1=4.3\mathrm{L}$

Since for an isobaric process we have $\frac{V}{T}=$const., the increase in temperature will be as many times as the increase in volume.

$T_2=3T_1=3·(273+20)=879\mathrm{K}=606°\mathrm{C}$

Final Answer (Part a)

That means that the temperature will be,

$T_2=3T_1=3·(273+20)=879\mathrm{K}=606°\mathrm{C}$

Given Information (Part b)

Amount of nitrogen gas$=5.0g$

Temperature$=20°\mathrm{C}$

Initial pressure$=3.0\mathrm{atm}$

Explanation (Part b)

(b) The heat exchanged in an isobaric process is given by

$Q=n{C}_p\Delta T$

Substituting the definition of $n$as previously and $\Delta T=3T_1-T_1=2T_1$we have

localid="1648620308631" $Q=\frac{2m}{M}{C}_p{T}_1=\left(\frac{2\times 0.005kg}{0.028kg/mol}\times 29.1Pa\times 293K\right)$

Multiply the values,

$=3050\mathrm{J}$

Final Answer (Part b)

Therefore, the amount of heat energy is transferred to the gas to cause this expansion is$3050J$.

Given Information (Part c)

Amount of nitrogen gas$=5.0\mathrm{g}$

Temperature$=20°\mathrm{C}$

Initial pressure$=3.0\mathrm{atm}$

Explanation (Part c)

(c) For an isochoric process, we have that the ratio between the pressure and the temperature remains constant. That means, that the pressure after the second process well be decreased by just as much as the temperature.

Since the temperature has to be decreased to the level prior to the first process, this means that we have a three- time decrease. That is,

$\frac{p_2}{T_2}=\frac{p_3}{T_3}=\frac{p_1}{T_2}=\frac{p_3}{T_1}\Rightarrow p_3=\frac{T_1}{T_2}{p}_1$

Knowing that the initial pressure was 3atm, we have

$p_3=1\mathrm{atm}$

Final Answer (Part c)

Hence, the gas pressure after the decrease is$1atm.$

Given Information (Part d)

Amount of nitrogen gas$=5.0\mathrm{g}$

Temperature$=20°\mathrm{C}$

Initial pressure$=3.0\mathrm{atm}$

Explanation (Part d)

(d) In an isochoric process the heat exchanged is given by

$Q=n{C}_v\Delta T$

Substituting just as in (b), with the difference that the temperature change here is negative, we have

localid="1648620401229" $Q=\frac{2m}{M}{C}_v{T}_1=\left(-\frac{2\times 0.005kg}{0.028kg/mol}\times 20.8m^3\times 293K\right)=-2180\mathrm{J}$

Final Answer (Part d)

Therefore, amount of heat energy is transferred from the gas as its pressure decreases is $-2180J.$The negative sign is an indicator that the heat was released by the gas.

Given Information (Part e)

Amount of nitrogen gas$=5.0\mathrm{g}$

Temperature$=20°\mathrm{C}$

Initial pressure$=3.0\mathrm{atm}$

Explanation (Part e)

The two processes are shown in a $p-V$graph below:

Final Answer (Part e)

Therefore, the diagram is as follows: