Respuesta :
Answer:
At -13 [tex]^{0}\textrm{C}[/tex] , the gas would occupy 1.30L at 210.0 kPa.
Explanation:
Let's assume the gas behaves ideally.
As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-
[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex]
where [tex]P_{1}[/tex] and [tex]P_{2}[/tex] are initial and final pressure respectively.
[tex]V_{1}[/tex] and [tex]V_{2}[/tex] are initial and final volume respectively.
[tex]T_{1}[/tex] and [tex]T_{2}[/tex] are initial and final temperature in kelvin scale respectively.
Here [tex]P_{1}=150.0kPa[/tex] , [tex]V_{1}=1.75L[/tex] , [tex]T_{1}=(273-23)K=250K[/tex], [tex]P_{2}=210.0kPa[/tex] and [tex]V_{2}=1.30L[/tex]
Hence [tex]T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}[/tex]
[tex]\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}[/tex]
[tex]\Rightarrow T_{2}=260K[/tex]
[tex]\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}[/tex]
So at -13 [tex]^{0}\textrm{C}[/tex] , the gas would occupy 1.30L at 210.0 kPa.
