Answer:
296.43 K is the temperature of a vessel when pressure reads 1.250 atm.
Explanation:
At STP, when pressure is 1.000 atm the value of temperature is 273,15 K.
If the pressure at temperature T reads as 1.250 atm.
[tex]P_1=1.000 atm, T_1=298.15K[/tex]
[tex]P_2=1.250 atm,T_2=?[/tex]
Applying Gay Lussac's law:
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex] (At constant volume)
[tex]\frac{1.000 atm}{273.15 K}=\frac{1.250 atm}{T_2}[/tex]
[tex]T_2=296.43 K[/tex]
296.43 K is the temperature of a vessel when pressure reads 1.250 atm.
The temperature of such a vessel, calibrated to read 1 atm at STP, if the pressure reads 1.250 atm is 341.25 K
The new temperature of the gas can be obtained as follow:
P₁ / T₁ = P₂ / T₂
1 / 273 = 1.25 / T₂
Cross multiply
T₂ = 273 × 1.25
T₂ = 341.25 K
Thus, the temperature will read 341.25 K
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