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Question: One mole of an ideal gas (\({{C}_{v}},m=\dfrac{5}{2}R\) ) at 300K and 5atm is expanded adiabatically...

One mole of an ideal gas (Cv,m=52R{{C}_{v}},m=\dfrac{5}{2}R ) at 300K and 5atm is expanded adiabatically to a final pressure of 2 atm against the constant pressure of 2 atm. The final temperature of the gas is:
(A) 270K
(B) 273K
(C) 248.5K
(D) 200K

Explanation

Solution

The state of the thermodynamic system evolves with slow time is a process of quasi-equilibrium. It is possible by doing work W on a system with increasing the internal energy U. if no heat is allowed in or out of a system during quasi-equilibrium is an adiabatic process.

Complete step by step solution:
In an adiabatic process, the system is insulated from its environment so that the state of the system changes, no heat is allowed to enter or leave from the system. This process can be conducted either quasi-statically or non-quasi-statically, when a system expands adiabatically, work done by the system and energy decreases which reflects the lowering of the temperature.
Similarly, when an ideal gas is compressed adiabatically (Q=0), work is done on the system by the surroundings.
From the first law of the thermodynamics, for an adiabatic process W=−PΔVW=-P\Delta V -- (1)
Given that,
Pext=2atm{{P}_{ext}}=2atm
Pressure change adiabatically, P1=5atm&P2=2atm{{P}_{1}}=5atm\And {{P}_{2}}=2atm
Temperature change, T1=300K&T2=?{{T}_{1}}=300K\And {{T}_{2}}=?
Given the number of moles = 1
Cv=52R{{C}_{v}}=\dfrac{5}{2}R
In the adiabatic process, W=nCv(T2−T1)W=n{{C}_{v}}({{T}_{2}}-{{T}_{1}}) --- (2)
From equation (1) and (2),
nCv(T2−T1)=−PΔVn{{C}_{v}}({{T}_{2}}-{{T}_{1}})=-P\Delta V
Substitute the given values in the above equation,
1X52R(T2−300)=−2(V2−V1)1X\dfrac{5}{2}R({{T}_{2}}-300)=-2({{V}_{2}}-{{V}_{1}})
52R(T2−300)=−2(nRT2P2−nRT1P1)=−2(1XRT22−1XRX3005)\dfrac{5}{2}R({{T}_{2}}-300)=-2(\dfrac{nR{{T}_{2}}}{{{P}_{2}}}-\dfrac{nR{{T}_{1}}}{{{P}_{1}}})=-2(\dfrac{1XR{{T}_{2}}}{2}-\dfrac{1XRX300}{5})
5T2−1500=−2T2+240 T2=248.57K \begin{aligned} & 5{{T}_{2}}-1500=-2{{T}_{2}}+240 \\\ & {{T}_{2}}=248.57K \\\ \end{aligned}
Therefore, the final temperature of the gas = 248.57K (approximately equal to 248.5K)

So, the right answer is option C.

Note: If ideal gas expands freely, the internal energy depends only on the temperature and there is no change in temperature during expansion. This is called the joule expansion. If this expansion is adiabatically then according to first law Q=0, W= 0, so internal energy is free for this expansion.