Reversible adiabatic compression

[eraymon2011]

Two moles of a diatomic gas occupies a volume of 50 L and is at a pressure 4atm. The gas then undergoes a process: a reversible adiabatic compression, so that is temperature doubles. Find both temperatures, and find the volume and the pressure after the compression process.

my thoughts: adiabatic compression- change in heat=dq= 0 for this process. I tried to find the intial temperature with the ideal gas law and got 1218 K (which seems a little high). I saw the formulas P2/P1= (V2/V1)^(-7/5) and T2= T1(P2/P1) ^(7/5-(1-7/5)) in a similar example (7/5= gamma for diatomic gas, the ^ symbol = exponent) and tried to apply them here but got P2 to be 11 atm and V2=11.45 L. I think that I'm wrong and I'm not sure where I went wrong.

 

[Nino MMP]

The initial temperature of the gas is given by the ideal gas law, $PV=nRT$. In this case, we have $4\text{atm}\cdot 50\text{L}=2\text{mol}\cdot R \cdot T_1$, so $T_1=\frac{4\cdot 50}{2\cdot R}\approx 1163K$.The volume and pressure of the gas after the reversible adiabatic compression can be found using the equation $P_2/P_1=(V_2/V_1)^{-7/5}$. Plugging in the known values for $P_1$ and $V_1$, we get $P_2=4\cdot (50/V_2)^{-7/5}\approx 11.2\text{atm}$. Then, we can solve for the volume after the compression process: $V_2=50\cdot (P_2/P_1)^{5/7}\approx 11.45\text{L}$.The temperature after the adiabatic compression can be found using the equation $T_2=T_1(P_2/P_1)^{7/5-(1-7/5)}=1163\cdot (11.2/4)^{2/5}\approx 2418K$.