Question

The temperature of 200.0 mL of a gas originally at STP is changed to -25°C at constant volume. Calculate the pressure of the gas in atm.

Answer 1

The pressure of the gas will be
`0.91atm`

Step-by-step explanation:

At STP, the temperature of the system is
`0^(o)C` while the pressure is
`1atm`

`T_1=0^oC=273.15K\\P_1=1atm`

The temperature is changed to
`-25^oC`, thus

`T_2=-25^oC=248.15K\\P_2=?`

As volume is constant i.e.,
`200.0ml`, by applying Gay-Lusaac law,

`(P_1)/(T_1) =(P_2)/(T_2)\\\\(1)/(273.15) =(P_2)/(248.15) \\\\P_2=(1)/(273.15) * 248.15 = 0.91atm`

Thus, the pressure of
`200.0ml` gas at
`-25^oC` is
`0.91atm`