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 The equation of state for a gas is given by  PV = nRT+\alpha V, where n is the number of moles and \alpha is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder are T_{0} and P_{0} respectively. The work done by the gas when its temperature doubles isobarically will be :

 

  • Option 1)

    \frac{P_{0}\: T_{0}\: R}{P_{0}-\alpha }

  • Option 2)

    \frac{P_{0}\: T_{0}\: R}{P_{0}+\alpha }

  • Option 3)

    P_{0}\: T_{0}\: R\: ln\: 2

  • Option 4)

    P_{0}\: T_{0}\: R

 

Answers (1)

best_answer

As we have learned

Bulk Modulus of elasticity in isobaric process -

\omega = \int p\: dV

=p\left ( V_{f}-V_{i} \right )
 

- wherein

W =p\left ( V_{f}-V_{i} \right )

=nR\left ( T_{f}-T_{i} \right )

 

 Equation of state is PV = nRT + dV 

\Rightarrow (P - \alpha ) V = nRT \: \: or \: \: P-\alpha = \frac{nRT}{V}

Initial volume = V_0 = \frac{nRT_0}{P_0- \alpha }

 

Final volume V_f = \frac{2nRT_0}{P_0- \alpha }

W = P\Delta V \\= P_0 (v_f -V_i ) = P_0 \left [ \frac{2nRT_0}{P_0 -\alpha} - \frac{nRT_0}{P_0-\alpha } \right ]

n = 1  w = p_0 \frac{RT_0}{P_0 - \alpha }

W = \frac{P_0T_0R}{P_0- \alpha }

 

 

 

 

 

 

 


Option 1)

\frac{P_{0}\: T_{0}\: R}{P_{0}-\alpha }

Option 2)

\frac{P_{0}\: T_{0}\: R}{P_{0}+\alpha }

Option 3)

P_{0}\: T_{0}\: R\: ln\: 2

Option 4)

P_{0}\: T_{0}\: R

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