Answer:
T'=92.70°C
Explanation:
To find the temperature of the gas you use the equation for ideal gases:
[tex]PV=nRT[/tex]
V: volume = 3000cm^3 = 3L
P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm
n: number of moles
R: ideal gas constant = 0.082 atm.L/mol.K
T: temperature = 27°C = 300.15K
For the given values you firs calculate the number n of moles:
[tex]n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles[/tex]
this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:
[tex]T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C[/tex]
hence, T'=92.70°C
