Answer:
The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.
Explanation:
When a container is rigid, the process is supposed to be isochoric, that is, at constant volume. Then, the equation of state for ideal gases can be simplified into the following expression:
![\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BP_%7B1%7D%7D%7BT_%7B1%7D%7D%20%3D%20%5Cfrac%7BP_%7B2%7D%7D%7BT_%7B2%7D%7D)
Where:
,
- Initial and final pressures, measured in pascals.
,
- Initial and final temperatures, measured in Kelvins.
In addtion, the specific heat at constant volume for monoatomic ideal gases, measured in joules per mole-Kelvin is given by:
![\bar c_{v} = \frac{3}{2}\cdot R_{u}](https://tex.z-dn.net/?f=%5Cbar%20c_%7Bv%7D%20%3D%20%5Cfrac%7B3%7D%7B2%7D%5Ccdot%20R_%7Bu%7D)
Where:
- Ideal gas constant, measured by pascal-cubic meters per mole-Kelvin.
If
, then:
![\bar c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{2}}{mol\cdot K} \right)](https://tex.z-dn.net/?f=%5Cbar%20c_%7Bv%7D%20%3D%20%5Cfrac%7B3%7D%7B2%7D%5Ccdot%20%5Cleft%288.314%5C%2C%5Cfrac%7BPa%5Ccdot%20m%5E%7B2%7D%7D%7Bmol%5Ccdot%20K%7D%20%5Cright%29)
![\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}](https://tex.z-dn.net/?f=%5Cbar%20c_%7Bv%7D%20%3D%2012.471%5C%2C%5Cfrac%7BJ%7D%7Bmol%5Ccdot%20K%7D)
And change in heat energy (
), measured by joules, by:
![Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})](https://tex.z-dn.net/?f=Q%20%3D%20n%5Ccdot%20%5Cbar%20c_%7Bv%7D%5Ccdot%20%28T_%7B2%7D-T_%7B1%7D%29)
Where:
- Molar quantity, measured in moles.
The final temperature of the monoatomic ideal gas is now cleared:
![T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%20T_%7B1%7D%20%2B%20%5Cfrac%7BQ%7D%7Bn%5Ccdot%20%5Cbar%20c_%7Bv%7D%7D)
Given that
,
,
and
, the final temperature is:
![T_{2} = 300\,K + \frac{6000\,J}{(4\,mol)\cdot \left(12.471\,\frac{J}{mol\cdot K} \right)}](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%20300%5C%2CK%20%2B%20%5Cfrac%7B6000%5C%2CJ%7D%7B%284%5C%2Cmol%29%5Ccdot%20%5Cleft%2812.471%5C%2C%5Cfrac%7BJ%7D%7Bmol%5Ccdot%20K%7D%20%5Cright%29%7D)
![T_{2} = 420.279\,K](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%20420.279%5C%2CK)
The final pressure of the system is calculated by the following relationship:
![P_{2} = \left(\frac{T_{2}}{T_{1}}\right) \cdot P_{1}](https://tex.z-dn.net/?f=P_%7B2%7D%20%3D%20%5Cleft%28%5Cfrac%7BT_%7B2%7D%7D%7BT_%7B1%7D%7D%5Cright%29%20%5Ccdot%20P_%7B1%7D)
If
,
and
, the final pressure is:
![P_{2} = \left(\frac{420.279\,K}{300\,K} \right)\cdot (6.00\times 10^{4}\,Pa)](https://tex.z-dn.net/?f=P_%7B2%7D%20%3D%20%5Cleft%28%5Cfrac%7B420.279%5C%2CK%7D%7B300%5C%2CK%7D%20%5Cright%29%5Ccdot%20%286.00%5Ctimes%2010%5E%7B4%7D%5C%2CPa%29)
![P_{2} = 8.406\times 10^{4}\,Pa](https://tex.z-dn.net/?f=P_%7B2%7D%20%3D%208.406%5Ctimes%2010%5E%7B4%7D%5C%2CPa)
The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.