Question

Please help! 10 points!!
What is the volume of gas at 2.00 atm and 200.0 K if its original volume was 300. L at STP?
109 .9 L
219.8 L
1.1 L
4.5 L

Answer 1

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:

$PV=nRT$

Or,

$(1 atm)(300 L)=nR(273.15 K)$

Or,

$nR=\frac{(1 atm)(300 L)}{(273.15 K)}=1.0983 atm L K^{-1}$

Now, volume at pressure 2 atm and T 200 K is to be calculated. Putting the temperature and pressure values in ideal gas equation,

$(2 atm)V=nR(200 K)$

Rearranging and putting the calculated value of nR,

$V=\frac{(1.0983 atm L K^{-1})(200 K)}{(2 atm)}=109.9 L$

Therefore, volume of gas at 2 atm and 200 K is 109.9 L that is option A.