A gas 2.48 L at 85.0 C and 725 torr. What is the volume will it occupy at STP ?
Ok well first combined law is P1xV1/ T1 = P2 xV2 /T2
Ok first I have to convert temp from C 273 + 85.0 = 358 K
Then I have to conver torr to atm 725 torr. x 1 Atm / 760 torr the torr cancel out and I have 0.9539 and make sure have sig fig. Making it 0.954
Now I plug all my number in
0.954x2.48 L / 358 K = 0.661 I guess atm
Ok well first combined law is P1xV1/ T1 = P2 xV2 /T2
Ok first I have to convert temp from C 273 + 85.0 = 358 K
Then I have to conver torr to atm 725 torr. x 1 Atm / 760 torr the torr cancel out and I have 0.9539 and make sure have sig fig. Making it 0.954
Now I plug all my number in
0.954x2.48 L / 358 K = 0.661 I guess atm
-
This is a combined gas law problem. We know that because you are making changes to the pressure, volume and temperature of a confined gas.
You do have to express temperature in an absolute temperature scale, like Kelvin. You DON'T have to change the pressure to atmospheres. The only thing you have to do is make sure that the pressures are in the same units and the volumes are in the same units. You can easily leave the pressure in torr. The only thing you have to know is the standard pressure in torr, which you seem to know is 760 torr. (Standard pressure: 1 atm, 760 torr, 760 mm Hg, 101.3 kPa...)
The second temperature and pressure are standard temperature and pressure, 273K and 760 torr.
P1V1/T1 = P2V2/T2
solve for V2
V2 = P1V1T2 / T1P2
V2 = 725 torr x 2.48L x 273K / 358K / 760 torr .... K and torr cancel out, leaving L
V2 = 1.80 L
============ Follow up =============
Again! You do *not* have to convert the pressure to atmospheres. Leave it in torr. It's must fine. All you do when you convert unnecessarily is run the risk of additional math errors and round-off errors. Your answer will have three significant digits because all three of your original values have three significant digits. Youliya's answer -- bless her heart, she's trying -- only has two significant digits.
You do have to express temperature in an absolute temperature scale, like Kelvin. You DON'T have to change the pressure to atmospheres. The only thing you have to do is make sure that the pressures are in the same units and the volumes are in the same units. You can easily leave the pressure in torr. The only thing you have to know is the standard pressure in torr, which you seem to know is 760 torr. (Standard pressure: 1 atm, 760 torr, 760 mm Hg, 101.3 kPa...)
The second temperature and pressure are standard temperature and pressure, 273K and 760 torr.
P1V1/T1 = P2V2/T2
solve for V2
V2 = P1V1T2 / T1P2
V2 = 725 torr x 2.48L x 273K / 358K / 760 torr .... K and torr cancel out, leaving L
V2 = 1.80 L
============ Follow up =============
Again! You do *not* have to convert the pressure to atmospheres. Leave it in torr. It's must fine. All you do when you convert unnecessarily is run the risk of additional math errors and round-off errors. Your answer will have three significant digits because all three of your original values have three significant digits. Youliya's answer -- bless her heart, she's trying -- only has two significant digits.
-
your answers is very close but not quit right
you are right about the Law P1*V1/T1=P2*V2/T2
YOU HAVE
P1= 0.954 atm
V1=2.48 L
T1= 358 K
so STP IS
P2= 1 atm
V2=?
T2= 273 k
so 0.954atm* 2.48L/358 K= 1 atm* P2/ 273 K so you cross multiply and P2
P2=.954 atm*2.48 L*273 K/358K= 1.80 L
you just forgot one of the piece's STP
GOOD LUCK
you are right about the Law P1*V1/T1=P2*V2/T2
YOU HAVE
P1= 0.954 atm
V1=2.48 L
T1= 358 K
so STP IS
P2= 1 atm
V2=?
T2= 273 k
so 0.954atm* 2.48L/358 K= 1 atm* P2/ 273 K so you cross multiply and P2
P2=.954 atm*2.48 L*273 K/358K= 1.80 L
you just forgot one of the piece's STP
GOOD LUCK
-
P1xV1/ T1 = P2 xV2 /T2
1 torr = 0.00131578947 atm
P1 = 725 * 0.00131578947 atm= 0.954 atm
temp 85 +273 = 358 K
0.954 * 2.48 /358 = 1* V2 /273
V2 = 1.8 L
the second Temp.& pressure as it u mention it in is at STP standared Temp & pressure wthich is 273 K & 1atm
1 torr = 0.00131578947 atm
P1 = 725 * 0.00131578947 atm= 0.954 atm
temp 85 +273 = 358 K
0.954 * 2.48 /358 = 1* V2 /273
V2 = 1.8 L
the second Temp.& pressure as it u mention it in is at STP standared Temp & pressure wthich is 273 K & 1atm