Pure helium is closed into cylinder with a movable piston. The initial volume, pressure, and temperature of the gas are 15 dm3, 2·105 Pa, and 300 K, respectively. If volume is decreased to 12 dm3 and the pressure increase to 3.5·105 Pa find the final temperature of the gas.
P V/T = n R
n and R are the same for both
P1 V1/T1 = P2 V2/T2
P1 V1 T2 = P2 V2 T1
T2 = T1 (P2/P1)(V2/V1)
luckily using degrees Kelvin so
T2 = 300 (3.5/2)(12/15)
To find the final temperature of the gas, we can use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
In this case, we have the initial conditions represented by subscripts "i" and the final conditions represented by subscripts "f":
P_i = 2·10^5 Pa
V_i = 15 dm^3
T_i = 300 K
P_f = 3.5·10^5 Pa
V_f = 12 dm^3
T_f = ? (what we want to find)
Since the number of moles is constant, it cancels out in the equation. So we can rewrite the equation as:
P_i * V_i / T_i = P_f * V_f / T_f
Now we can rearrange the equation to solve for T_f:
T_f = (P_f * V_f * T_i) / (P_i * V_i)
Substituting the given values:
T_f = (3.5·10^5 Pa * 12 dm^3 * 300 K) / (2·10^5 Pa * 15 dm^3)
Now we can calculate the final temperature:
T_f = 2520 K
Therefore, the final temperature of the gas is 2520 K.