Respuesta :
For the above question we can use the combined gas law,
[tex] \frac{P1V1}{T1} = \frac{P2V2}{T2} [/tex]
where P - pressure, T- temperature and V- volume
temperature has been given in celsius, we have to convert into Kelvin scale to calculate.
temperature in Kelvin = temperature in celcius + 273
27 °C + 273 = 300 K
[tex]\frac{1.05 atm*121 mL}{300K} = \frac{1.40 atm*293 mL }{T} [/tex]
T = 967 K
Temperature in celcius - 969 - 273 = 696 °C
[tex] \frac{P1V1}{T1} = \frac{P2V2}{T2} [/tex]
where P - pressure, T- temperature and V- volume
temperature has been given in celsius, we have to convert into Kelvin scale to calculate.
temperature in Kelvin = temperature in celcius + 273
27 °C + 273 = 300 K
[tex]\frac{1.05 atm*121 mL}{300K} = \frac{1.40 atm*293 mL }{T} [/tex]
T = 967 K
Temperature in celcius - 969 - 273 = 696 °C
Answer: The final temperature is coming out to be 704.9°C
Explanation:
To calculate the temperature when pressure and volume has changed, we use the equation given by combined gas law. The equation follows:
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1,V_1\text{ and }T_1[/tex] are the initial pressure, volume and temperature of the gas
[tex]P_2,V_2\text{ and }T_2[/tex] are the final pressure, volume and temperature of the gas
We are given:
[tex]P_1=1.05atm\\V_1=121mL\\T_1=27^oC=[27+273]K=300K\\P_2=1.40atm\\V_2=293mL\\T_2=?K[/tex]
Putting values in above equation, we get:
[tex]\frac{1.04atm\times 121mL}{300K}=\frac{1.40atm\times 293mL}{T_2}\\\\T_2=\frac{1.40\times 293\times 300}{1.04\times 121}=977.9K[/tex]
Converting the temperature from kelvins to degree Celsius, by using the conversion factor:
[tex]T(K)=T(^oC)+273[/tex]
[tex]977.9=T(^oC)+273\\T(^oC)=704.9^oC[/tex]
Hence, the final temperature is coming out to be 704.9°C
