2.94
×
10
24
⋅
molecules
6.022
×
10
23
⋅
molecules
⋅
mol
−
1
×
142.3
⋅
g
⋅
mol
−
1
≅
700
⋅
g
The top one does because there are more and it’s closer
<u>Answer:</u> The new volume of the gas is 0.11 L
<u>Explanation:</u>
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.
The equation follows:
![\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
where,
are the initial pressure, volume and temperature of the gas
are the final pressure, volume and temperature of the gas
At STP:
The temperature at this condition is taken as 273 K and the pressure at this condition is taken as 1 atm or 101.3 kPa.
We are given:
![P_1=101.3kPa\\V_1=0.2L\\T_1=273K\\P_2=202.6kPa\\V_2=?\\T_2=300K](https://tex.z-dn.net/?f=P_1%3D101.3kPa%5C%5CV_1%3D0.2L%5C%5CT_1%3D273K%5C%5CP_2%3D202.6kPa%5C%5CV_2%3D%3F%5C%5CT_2%3D300K)
Putting values in above equation, we get:
![\frac{101.3kPa\times 0.2L}{273K}=\frac{202.6kPa\times V_2}{300K}\\\\V_2=\frac{101.3\times 0.2\times 300}{273\times 202.6}=0.11L](https://tex.z-dn.net/?f=%5Cfrac%7B101.3kPa%5Ctimes%200.2L%7D%7B273K%7D%3D%5Cfrac%7B202.6kPa%5Ctimes%20V_2%7D%7B300K%7D%5C%5C%5C%5CV_2%3D%5Cfrac%7B101.3%5Ctimes%200.2%5Ctimes%20300%7D%7B273%5Ctimes%20202.6%7D%3D0.11L)
Hence, the new volume of the gas is 0.11 L
I think it’s b I’m not sure so sorry if incorrect. Take care.