Option A: 109.9 L
At STP, pressure and temperature of gas is 1 atm and 273.15 K respectively.
It is given that volume at STP is 300 L, putting all the values in ideal gas equation as follows:
![PV=nRT](https://img.qammunity.org/2019/formulas/chemistry/college/sadva412qmvxq3yijfpjgu3po09xu9zppw.png)
Or,
![(1 atm)(300 L)=nR(273.15 K)](https://img.qammunity.org/2019/formulas/chemistry/high-school/25s6axkdmjps740yjurzke931pf33tjl4i.png)
Or,
![nR=((1 atm)(300 L))/((273.15 K))=1.0983 atm L K^(-1)](https://img.qammunity.org/2019/formulas/chemistry/high-school/ob9fbw5ldte7hysmy3edztwyl14b5u56cu.png)
Now, volume at pressure 2 atm and T 200 K is to be calculated. Putting the temperature and pressure values in ideal gas equation,
![(2 atm)V=nR(200 K)](https://img.qammunity.org/2019/formulas/chemistry/high-school/s8bcjhw4ds8uygkh7eywynhcxqyjw98e0w.png)
Rearranging and putting the calculated value of nR,
![V=((1.0983 atm L K^(-1))(200 K))/((2 atm))=109.9 L](https://img.qammunity.org/2019/formulas/chemistry/high-school/s7pulf6ln6eg4452rre79sbbfxwbbkahbf.png)
Therefore, volume of gas at 2 atm and 200 K is 109.9 L that is option A.