Question

One mole of an ideal gas undergoes a process
p=p01+(V/V0)2
where p0 and V0 are constants. Find the temperature of the gas when V = V0.

Answer 1

One mole of an ideal gas undergoes a process $P=\dfrac{P_0}{1+(\frac{V}{V_0})^2}$ where P0 and V0  are constants, the temperature of the gas when $V=V_0$ is

$T=\dfrac{P_0V_0}{2R}$.

• Given,
• $P=\dfrac{P_0}{1+(\frac{V}{V_0})^2}$
• according to the ideal gas equation, we have,
• PV=nRT
• ⇒
• $\dfrac{nRT}{V}=\dfrac{P_0}{1+(\frac{V}{V_0})^2}$
• ⇒
• Given, n=1 mole, we have,
• $\dfrac{RT}{V}=\dfrac{P_0}{1+(\frac{V}{V_0})^2}$
• at $V=V_0$
• $\dfrac{RT}{V_0}=\dfrac{P_0}{1+(\frac{V_0}{V_0})^2}\\\\\dfrac{RT}{V_0}=\dfrac{P_0}{1+1}$
• ⇒$P_0V_0=RT(1+1)\\\\P_0V_0=2RT$

• ∴$T=\dfrac{P_0V_0}{2R}$

Answer 2

The temperature of the gas $T=\frac{P_{0} V_{0}}{2 R}$

Explanation:

Given data in the question  

One ideal gas mole undergoes a process $\mathrm{P}=\frac{P_{0}}{1+\left(\frac{V}{V_{0}}\right)^{2}}$

where $p_0 \text {and} v_0$  are constants .  

$\mathrm{P}=\frac{P_{0}}{1+\left(\frac{V}{V_{0}}\right)^{2}}$

With both sides multiplied by V in the above eqation we get

$\mathrm{PV}=\frac{P_{0}V}{1+\left(\frac{V}{V_{0}}\right)^{2}}$  . . . . . . . . . . . . . .  equation ( 1 )

We know that  

PV=nRT  

where n = 1

So PV=RT

Put the value of PV=RT in equation ( 1 )

Now we get

$R T=\frac{P_{o} V}{1+\left(\frac{V}{V_{o}}\right)^{2}}$

$T=\frac{1}{R} \frac{P_{o} V}{1+\left(\frac{V}{V_{o}}\right)^{2}}$

V = Vo

$T=\frac{1}{R} \frac{P_{o} V_{o}}{1+\left(\frac{V_{o}}{V_{o}}\right)^{2}}$

$T=\frac{1}{R} \frac{P_{o} V_{o}}{\left(1+1^{2}\right)}$

$T=\frac{1}{R} \frac{P_{o} V_{o}}{2}$

$T=\frac{P_{0} V_{0}}{2 R}$

Therefore for one more of an ideal gas undergoes a process  $\mathrm{P}=\frac{P_{0}}{1+\left(\frac{V}{V_{0}}\right)^{2}}$  where  $p_0 \text {and} v_0$ are constants and the temperature of the gas is $T=\frac{P_{0} V_{0}}{2 R}$ with condition V = $V_0$