Volume stays constant, so: P1/T1 = P2/T2
P1 = 3 atm
T1 = 20*C = 293 K
P2 = ?
T2 = 50*C = 323 K
Solve for P2: P1T2/T1 = P2
P2 = (3 atm)(323 K)/(293 K) = 3.3 atm
The Pressure of N2 after its temperature increased will be "3.3 atm". To understand the calculation, check below.
According to the question,
Pressure, P₁ = 3 atm
Temperature, T₁ = 20°C or,
= 293 K
T₂ = 50°C or,
= 323 K
We know the relation,
→ [tex]\frac{P_1}{T_1}[/tex] = [tex]\frac{P_2}{T_2}[/tex]
or,
Pressure, P₂ = [tex]\frac{P_1\times T_2}{T_1}[/tex]
By substituting the values,
= [tex]\frac{3\times 323}{293}[/tex]
= [tex]\frac{969}{293}[/tex]
= 3.3 atm
Thus the above response is correct.
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