According to Gay -Lussac’s law (at constant volume, as pressure increases, temperature increases)
[tex]\frac{P_{1} }{T_{1} } =\frac{P_{2} }{ T_{2} }[/tex]
Given, [tex]P_{1} =1.00\times 10^{3} \ kPa[/tex] , [tex]T_{1} =19^o C = (19 +273 ) = 292 \ k[/tex] and [tex]T_{2} =46^o C = (46 +273 ) = 319 \ k[/tex]
Substituting these values, in above formula, we get
[tex]\frac{1.00\times 10^{3} \ kPa}{292 \ K} = \frac{P_{2} }{319 \ K}\\\\ P_{2} =\frac{1.00\times 10^{3} \ kPa \times 319 K}{292K} = 1.09 \times 10^3 \ kPa[/tex].
Thus, the new pressure in the cylinder is [tex]1.09 \times 10^3 \ kPa[/tex].
