A certain volume of a gas at 298kelvin is heated such that its volume and pressure are now four times their original value.What is the new tewperature?
Use (P1V1/T1) = (P2V2/T2)
The easy way to do this is to make up a convenient number for P1 and V1 and multiply by 4 to find P2 and V2.
Another way is to call P1 = P1, then P2 = 4P1.
While V1 = V1 and V2 = 4V1
V1-40
V2-160
P1-90
P2-360
T1-298K
T2-?
V1P1/T1=V2P2/T2
To find the new temperature (T2), you can use the combined gas law equation. The combined gas law equation is derived from the ideal gas law and it relates the initial and final states of pressure, volume, and temperature.
The combined gas law equation is given as follows:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature (298 K)
P2 = final pressure (4 times the initial pressure)
V2 = final volume (4 times the initial volume)
T2 = final temperature (unknown)
From the information given in the question, the final volume (V2) and pressure (P2) are four times their original values. Therefore, we have:
P2 = 4 * P1
V2 = 4 * V1
Now, we can substitute these values into the combined gas law equation:
(P1 * V1) / T1 = (4 * P1 * 4 * V1) / T2
Simplifying the equation further:
(P1 * V1 * T2) = (4 * P1 * 4 * V1 * T1)
Canceling out the common terms (P1 and V1):
T2 = (4 * 4 * T1) = 16 * T1
Substituting the initial temperature value (T1 = 298 K) into the equation:
T2 = 16 * 298 K
Calculating the value:
T2 = 4768 K
Therefore, the new temperature (T2) after heating the gas is 4768 Kelvin.