Respuesta :
Answer:
The final temperature is 503°C.
Explanation:
Given that,
Volume = 499 cm³
Gauge pressure [tex]P_{i}= 1.01\times10^{5}\ Pa[/tex]
Temperature = 27.0°
Final volume = 46.2 cm³
Increased pressure [tex]P_{f}= 2.72\times10^{6}\ Pa[/tex]
We need to calculate the temperature
Using ideal gas equation
[tex]PV=nRT[/tex]
Rearrange equation
[tex]T_{2}=\dfrac{P_{2}V_{2}}{P_{1}V_{1}}\timesT_{1}[/tex]
Put the value into the formula
[tex]T_{2}=\dfrac{(2.72\times10^{6}+1.01\times10^{5})\times46.2\times10^{-6}}{1.01\times10^{5}\times499\times10^{-6}}\times(27+273)[/tex]
[tex]T_{2}=776\ K[/tex]
[tex]T_{2}=776-273=503\ ^{\circ}C[/tex]
Hence, The final temperature is 503°C.
The final temperature is 502.8 degrees C.
How do you calculate the final temperature?
Given that the volume V1 of the cylinder is 499 cm³, the gauge pressure P1 is 1.01×10^5 Pa, temperature t1 is 27.0°.
The compressed volume V2 of the cylinder is 46.2 cm³ and the increased pressure is 2.72×10^6 Pa.
Then the final pressure is P2 = 2.72×10^6 Pa + 1.01×10^5 Pa.
As per the ideal gas equation,
[tex]PV = nRT[/tex]
For the initial condition, the gas equation is,
[tex]P_1V_1 = nRT_1[/tex]
[tex]nR = \dfrac {P_1V_1}{T_1}[/tex]............equation 1
For the final condition, the gas equation is,
[tex]P_2V_2=nRT_2[/tex]
[tex]nR = \dfrac {P_2V_2}{T_2}[/tex]...........equation 2
When equating both the equations, 1 and 2, we get,
[tex]\dfrac {P_1V_1}{T_1}=\dfrac {P_2V_2}{T_2}[/tex]
[tex]T_2 = \dfrac {P_2V_2}{P_1V_1}\tiems T_1[/tex]
Substituting the values, we get,
[tex]T_2 = \dfrac {(2.72\times 10^6+1.01\times 10^5)\times 46.2\times 10^{-6}}{1.01\times 10^5\times 499\times 10^{-6}}\times (27+273)[/tex]
[tex]T_2 = 775.8\;\rm K[/tex]
[tex]T_2 = 775.8-273 = 502.8^\circ\;\rm C[/tex]
Hence we can conclude that the final temperature is 502.8 degrees C.
To know more about the ideal gas law, follow the link given below.
https://brainly.com/question/4147359.
