If 10.0 L of a gas exerts 5.66 atm at 20 degrees Celsius, what would the temperature be when gas exerted 18.7 atm? Your answer must include the correct units.
To find the new temperature when the gas exerts 18.7 atm, we can use the combined gas law formula:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = 5.66 atm
V1 = 10.0 L
T1 = 20 degrees Celsius = 293 K
P2 = 18.7 atm
V2 = 10.0 L
T2 = unknown
Plugging in the values:
(5.66 atm * 10.0 L) / 293 K = (18.7 atm * 10.0 L) / T2
T2 = (18.7 atm * 10.0 L * 293 K) / (5.66 atm * 10.0 L)
T2 = 9724.10 K
Therefore, the temperature would be 9724.10 K.
If 10.0 L of a gas exerts 5.66 atm at 20 degrees Celsius, what would the temperature be when gas exerted 18.7 atm? Your answer must include the correct units.
According to the question above, if instead we kept the temperature constant, what u pressure be if the volume was doubled?
To find the pressure when the volume is doubled while keeping the temperature constant, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when temperature is held constant.
Boyle's Law formula:
P1 * V1 = P2 * V2
Given:
P1 = 5.66 atm
V1 = 10.0 L
P2 = unknown
V2 = 2 * V1 = 20.0 L
Now we can plug in the values to solve for P2:
5.66 atm * 10.0 L = P2 * 20.0 L
56.6 atm·L = 20P2
Divide by 20 on both sides to solve for P2:
P2 = 56.6 atm·L / 20
P2 = 2.83 atm
Therefore, if the volume of the gas is doubled while keeping the temperature constant, the pressure would be 2.83 atm.