A mass of gas at 40mmHg is heated from 298k to 348k at constant volume. Cal the pressure exerted by the gas.
Answer Details
To determine the new pressure exerted by the gas when it is heated, we'll apply **Gay-Lussac's Law**. This law states that at constant volume, the pressure of a given amount of gas is directly proportional to its absolute temperature. Mathematically, it can be expressed as:
P1/T1 = P2/T2
Where:
P1 = Initial pressure of the gas (40 mmHg)
T1 = Initial temperature (298 K)
P2 = Final pressure of the gas, which we want to find
T2 = Final temperature (348 K)
By rearranging the formula to solve for the final pressure (P2), we get:
P2 = P1 * (T2/T1)
Now, insert the given values into the equation:
P2 = 40 mmHg * (348 K / 298 K)
Perform the calculations:
P2 = 40 mmHg * (348 / 298)
P2 = 40 mmHg * 1.1678
P2 = 46.71 mmHg
So, the new pressure exerted by the gas when it is heated from 298 K to 348 K at constant volume is 46.71 mmHg.