Question

Temperature of 1 mole of an ideal gas is
increased from 300 K to 310 Kunder
isochoric process. Heat supplied to the gas
in this process is 2 = 25 R, where R=universal
gas constant. The amount of work that has
to be done by the gas if temperature of the
gas decreases from 310 K to 300 K
adiabatically is​

Answer 1

Given:

Temperature of 1 mole of an ideal gas is increased from 300 K to 310 K under isochoric process. Heat supplied to the gas in this process is Q = 25 R, where R=universal gas constant.

To find:

The amount of work that has to be done by the gas if temperature of the gas decreases from 310 K to 300 K in adiabatic process.

Calculation:

Applying 1st Law of Thermodynamics in Isochoric process:

$\rm{\Delta \: Q = \Delta \: U + W}$

$= > \rm{\Delta \: Q = \Delta \: U + 0}$

$= > \rm{\Delta \: Q = \Delta \: U }$

$= > \rm{ \Delta \: U = + 25R }$

$= > \rm{ nC_{V}\Delta\theta = + 25R }$

$= > \rm{ nC_{V}(310-300) = + 25R }$

$= > \rm{ nC_{V}(10) = + 25R }$

$= > \rm{ nC_{V} = 2.5R }$

Now, again applying 1st Law of thermodynamics in the adiabatic process:

$\rm{\Delta \: Q = \Delta \: U + W}$

$= > \rm{0 = \Delta \: U + W}$

$= > \rm{ \Delta \: U = - W}$

$= > \rm{W = - \Delta\: U }$

$= > \rm{W = - nC_{V}\Delta\theta }$

$= > \rm{W = - (2.5R)(300-310) }$

$= > \rm{W = - (2.5R)(-10) }$

$= > \rm{W = 25R }$

So final answer is:

$\boxed{\bf{Work\:done = 25R}}$

Answer 2

Answer:

The amount of work done by the gas in adiabatic process is 25R.

Explanation:

Given 1 mole of an ideal gas, $n=1$

Initial temperature, $T_{i} =300 K$

Final temperature, $T_{f} =310 K$ to 310

Heat energy supplied to the gas is $Q = 25 R$, where R is the universal

gas constant.

In isochoric process, volume is constant.

Therefore, from the first law of thermodynamics,

$\Delta Q=\Delta U+W$

For an ideal gas, the internal energy is function of temperature and is given by $\Delta U =nC_{V} \Delta T$ and work done is given by $W =PdV$.

Therefore, $\Delta Q=nC_{V} \Delta T+PdV$

Substituting the values,

$25R=1\times C_{V} (310-300)$

$\implies 25R=C_{V} (10)\implies C_{V}=\frac{25R}{10} =2.5R$

In adiabatic process, no heat energy is transferred between a system and its surroundings, i.e.,  $\Delta Q=0$.

Therefore, from the first law of thermodynamics, $\Delta Q=nC_{V} \Delta T+W$

$\implies0=nC_{V} \Delta T+W \implies W=-nC_{V} \Delta T$

Substituting the values, when temperature is decreased from 310 K to 300 K is

$W=-1\times 2.5R\times (300-310)$    

$=-2.5R\times(-10)=25R$

Therefore, the amount of work done by the gas is 25R.