Answer:
![15.5^0C](https://img.qammunity.org/2021/formulas/chemistry/high-school/z7wte1z9w56w9ojo3v4x4i7bvctn7q0ayg.png)
Step-by-step explanation:
According to ideal gas equation:
![PV=nRT](https://img.qammunity.org/2021/formulas/physics/high-school/xmnfk8eqj9erqqq8x8idv0qbm03vnipq7i.png)
P = pressure of gas = 0.987 atm
V = Volume of gas = 12 L
n = number of moles = 0.50
R = gas constant =
![0.0821Latm/Kmol](https://img.qammunity.org/2021/formulas/chemistry/college/lrfckhcxrz16jyka569n65jplk2jrnikrv.png)
T =temperature = ?
![T=(PV)/(nR)](https://img.qammunity.org/2021/formulas/chemistry/high-school/ywv32bgmkwg8o6v9loufx5vtqjntav4xt4.png)
![T=(0.987atm* 12L)/(0.0820 L atm/K mol* 0.50mol)=288.5K=(288.5-273)^0C=15.5^0C](https://img.qammunity.org/2021/formulas/chemistry/high-school/y122pfjbqojadqhjsh0xep05nlqfb5wndg.png)
Thus the temperature of a 0.50 mol sample of a gas at 0.987 atm and a volume of 12 L is
![15.5^0C](https://img.qammunity.org/2021/formulas/chemistry/high-school/z7wte1z9w56w9ojo3v4x4i7bvctn7q0ayg.png)