91Ó°ÊÓ

Nitrogen gas at STP $=14g$

Pressure $=20atm$

Final temperature$=?$

(a) Since this process is isobaric, we have that the ratio between the pressure and temperature remains constant. That is,

$\frac{p_1}{T_1}=\frac{p_2}{T_2}$

We can therefore express the temperature as

$T_2=\frac{p_2}{p_1}{T}_1$

Knowing the initial temperature and - having assumed that $100\mathrm{kPa}$is equivalent to $1atm$- knowing that the pressure ratio is 20 , we can find

$T_2=20·273=5460\mathrm{K}$

Hence, the final temperature is $5460K.$

Nitrogen gas at STP $=14g$

Pressure$=20\mathrm{atm}$

The work done on the gas $=?$

There is no volume change in this process, because it is isochoric. There is no work done because there is no volume change.

Hence, the work done on the gas is Zero.

Nitrogen gas at STP$=14\mathrm{g}$

Pressure$=20\mathrm{atm}$

The heat transfer to the gas$=?$

(c) The heat given in an isochoric process is given by

$Q=n{C}_v\mathrm{Δ}T$

Knowing that $T_2=20T_1$, we have that the temperature difference is $\mathrm{Δ}T=19T_1=19·273=5187\mathrm{K}$.

We also need to find the number of moles, knowing that we have a mass of $m=14\mathrm{gr}$and that the molar mass of nitrogen gas is $M=28\mathrm{gr}/\mathrm{mol}$, we have

$n=\frac{m}{M}=\frac{14}{28}=0.5$

Substituting the above mentioned, we have

$Q=0.5·20.8·5187=53950\mathrm{J}$

Hence, the heat transfer to the gas is$53950J.$

Nitrogen gas at STP $=14\mathrm{g}$

Pressure$=20\mathrm{atm}$

The pressure ratio$=?$

(d) As mentioned before, the ratio between the maximum and the minimum pressures is 20. This is having rounded the atmospheric pressure as $101,300\mathrm{Pa}≈100,000\mathrm{Pa}$.

However, if we want to be exact and find this precise ratio, we will get

$\frac{p_2}{p_1}=\frac{20·1.013·{10}^5}{1·{10}^5}=20.26$

Hence, the pressure ratio is $P_{\max }/P_{\min }=20$or $20.6$

Nitrogen gas at STP$=14\mathrm{g}$

Pressure$=20\mathrm{atm}$

A graphical representation of the process is given below.

The process on a $pV$diagram is,