The pressure will be -110 Kpa when the temperature is increased to 200.0 °C.
The pressure of the gas is directly proportional to the temperature of the gas.
Gay-lussac law
[tex]\dfrac{P_1}{T_1} =\dfrac{P_2}{T_2}\\\\\\\dfrac{55}{-100.0} =\dfrac{P_2}{200.0} \\\\P_2 = -110 Kpa[/tex]
Thus, the pressure will be -110 Kpa.
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