
One mole of an ideal diatomic gas at 300K having 1atm pressure is first heated to 400K at 1 atm pressure and then expanded reversibly to 0.5atm pressure at 400K. The total work done on the gas in terms of the universal gas constant R is
A. -100R(1+4ln2)
B. 100R(-1+4ln2)
C. 100R(1+4ln2)
D. 100R(1-4ln2)
Answer
497.7k+ views
Hint: The mole is considered as the basic unit for measuring or expressing the amount of the substance. It is defined as the amount of substance which is equal to the substance present in 12g of carbon 12 atom. We can calculate the moles of the substance by dividing the mass of the substance by the molar mass.
Complete step by step answer:
Let us first see the given data.
Given gas is a diatomic gas.
Let the initial temperature be $${T_1}$$ = 300K
Let the initial pressure be $${P_1}$$ = 1atm
Let the final temperature be $${T_2}$$ = 400K
Let the final pressure be $${P_2}$$ = 0.5atm
From the data given in question, it is observed that initially the pressure is constant and temperature changes. Hence it expresses an isobaric system. Later at the final stage, temperature is constant and pressure changes. Hence this expresses that the system is isothermal.
So, to calculate the total work done for gas to expand reversibly, we add the work done by gas both in isobaric condition and isothermal conditions.
W = ${W_{isobaric}}$ + ${W_{isothermal}}$
W = nR $({T_2} - {T_1})$ + nRT ln $\left[ {\dfrac{{{V_2}}}{{{V_1}}}} \right]$
W = nR $({T_2} - {T_1})$ + nRT ln $\left[ {\dfrac{{{P_1}}}{{{P_2}}}} \right]$, Since for ideal gas ${P_1}{V_1} = {P_2}{V_2}$ $ \Rightarrow $$\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{V_2}}}{{{V_1}}}$
On substituting the values, we get,
W = nR (400-300) + nR 400 ln$\left[ {\frac{{1.0}}{{0.5}}} \right]$
Since n = 1mole
W = 1 $ \times $ R $ \times $ (400-300) + 1 $ \times $R $ \times $ 400 $ \times $ ln$\left[ {\dfrac{{1.0}}{{0.5}}} \right]$
W = R (400-300) + R 400 ln$\left[ {\dfrac{{1.0}}{{0.5}}} \right]$
W = 100R + 400R ln2
On taking 100R common on the right-hand side,
W = 100R (1+ 4ln2)
So, the correct answer is Option C.
Note: The constant for gas is a physical constant which has been expressed in terms of units of energy per temperature increment of per mole. The other name for gas constant is molar gas constant or the universal gas constant. This gas constant is equal to the value of the Boltzmann constant which is expressed as the volume pressure product.
Complete step by step answer:
Let us first see the given data.
Given gas is a diatomic gas.
Let the initial temperature be $${T_1}$$ = 300K
Let the initial pressure be $${P_1}$$ = 1atm
Let the final temperature be $${T_2}$$ = 400K
Let the final pressure be $${P_2}$$ = 0.5atm
From the data given in question, it is observed that initially the pressure is constant and temperature changes. Hence it expresses an isobaric system. Later at the final stage, temperature is constant and pressure changes. Hence this expresses that the system is isothermal.
So, to calculate the total work done for gas to expand reversibly, we add the work done by gas both in isobaric condition and isothermal conditions.
W = ${W_{isobaric}}$ + ${W_{isothermal}}$
W = nR $({T_2} - {T_1})$ + nRT ln $\left[ {\dfrac{{{V_2}}}{{{V_1}}}} \right]$
W = nR $({T_2} - {T_1})$ + nRT ln $\left[ {\dfrac{{{P_1}}}{{{P_2}}}} \right]$, Since for ideal gas ${P_1}{V_1} = {P_2}{V_2}$ $ \Rightarrow $$\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{V_2}}}{{{V_1}}}$
On substituting the values, we get,
W = nR (400-300) + nR 400 ln$\left[ {\frac{{1.0}}{{0.5}}} \right]$
Since n = 1mole
W = 1 $ \times $ R $ \times $ (400-300) + 1 $ \times $R $ \times $ 400 $ \times $ ln$\left[ {\dfrac{{1.0}}{{0.5}}} \right]$
W = R (400-300) + R 400 ln$\left[ {\dfrac{{1.0}}{{0.5}}} \right]$
W = 100R + 400R ln2
On taking 100R common on the right-hand side,
W = 100R (1+ 4ln2)
So, the correct answer is Option C.
Note: The constant for gas is a physical constant which has been expressed in terms of units of energy per temperature increment of per mole. The other name for gas constant is molar gas constant or the universal gas constant. This gas constant is equal to the value of the Boltzmann constant which is expressed as the volume pressure product.
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