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# A sample of a gas has a pressure of 3.00 atm at 25°c. what would the gas pressure be at 52°c, if the volume remains constant?

The problem describes Gay-Lussac's Law in which pressure of the gas is directly proportional to its absolute temperature at constant volume. From Ideal gas equation, Gay-Lussac's Law then expressed as P=kT where k is the proportionality constant. Given:  P_1 = 3.00 atm  T_1 = 25°C (298.15 K)  T_2 = 52°C (325.15 K) Required:  P_2 = ? Assumptions:  Sample of gas is an ideal gas  Constant volume Solution: We say subscript 1 means first state and subscript 2 means second state  P=kT k = P/T k_1 = k_2    since it is the proportionality constant so k in first state is equal to k to second state Therefore,  $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ Substitute all known values to the equation You will get P_2 = [(3.00 atm)(325.15K)]/298.15K ∴ P_2 = 3.27 atm