A sample of gas of unknown pressure occupies 0.766 L at a temperature of 298 K. The same sample of gas is then tested under known conditions and has a pressure of 32.6 kPa and occupies 0.644 L at 303 K. What was the original pressure of the gas

26.96 kPa

Explanation:

Applying,

PV/T = P'V'/T' Equation 1

Where P = Initial pressure, V = Initial volume, T = Initial temperature, P' = Final pressure, V' = Final volume, T' = Final temperature.

make P the subject of the equation

P = P'V'T/VT' Equation 2

From the question,

Given: V = 0.766 L, T = 298 K, P' = 32.6 kPa, V' = 0.644 L, T' = 303 K

Substitute these values into equation 2

P = (32.6×0.644×298)/(0.766×303)

P = 26.96 kPa

the answer would be a