
An ideal gas initially at 300K undergoes an isobaric expansion at 2.50kPa. If the volume increases from 1.00${m^3}$ to 3.00${m^3}$and 12.5kJ is transferred to the gas by heat; what is its final temperature?
(A) 900K
(B) 400K
(C) 500K
(D) 200K
Answer
584.7k+ views
Hint- Isobaric expansion occurs at constant pressure. Work done is given by $W = P\Delta V$. Also, we have the Charles law$\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$ .
Complete step by step solution:
Following information is given in the question: ${T_1}$=300K, ${V_1} = $ 1 ${m^3}$ and ${V_2}$= 3${m^3}$.
Now, let us use the formula $\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$to calculate the final temperature.
Let us substitute the values.
$\dfrac{1}{{300}} = \dfrac{3}{{{T_2}}}$
On simplifying the expression, we get the following.
${T_2} = 900K$
Hence, option (A) 900K is the correct option.
Additional information:
*An ideal gas is that whose molecules occupy negligible space and its molecules do not interact with each other.
*Ideal gas obeys the ideal gas law always.
*PV=nRT, this is the ideal gas equation. where, P is pressure of the gas, V is volume of the, T is temperature of the given gas, n is number of moles that is present in the gas and R is universal gas constant.
*In isobaric processes pressure is kept constant. Example of an isobaric process is reversible expansion of the ideal gas.
*The molecules of gas are made up of small infinite particles and molecules of same gases differ from the molecules of other gases
*The collisions of the molecules are perfectly elastic and the kinetic energy between them is conserved.
*All the molecules of gas travel in the straight line unless they collide with the walls.
Note:
*Gas pressure is much less than the atmospheric pressure.
*This could be managed by keeping the gas in a cylinder with a piston on the bottom supporting a constant load that hangs from the piston.
*Now, the cylinder is put into the warm environment to expend the gas.
*Using the ideal gas equation, we can find the temperature.
Complete step by step solution:
Following information is given in the question: ${T_1}$=300K, ${V_1} = $ 1 ${m^3}$ and ${V_2}$= 3${m^3}$.
Now, let us use the formula $\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$to calculate the final temperature.
Let us substitute the values.
$\dfrac{1}{{300}} = \dfrac{3}{{{T_2}}}$
On simplifying the expression, we get the following.
${T_2} = 900K$
Hence, option (A) 900K is the correct option.
Additional information:
*An ideal gas is that whose molecules occupy negligible space and its molecules do not interact with each other.
*Ideal gas obeys the ideal gas law always.
*PV=nRT, this is the ideal gas equation. where, P is pressure of the gas, V is volume of the, T is temperature of the given gas, n is number of moles that is present in the gas and R is universal gas constant.
*In isobaric processes pressure is kept constant. Example of an isobaric process is reversible expansion of the ideal gas.
*The molecules of gas are made up of small infinite particles and molecules of same gases differ from the molecules of other gases
*The collisions of the molecules are perfectly elastic and the kinetic energy between them is conserved.
*All the molecules of gas travel in the straight line unless they collide with the walls.
Note:
*Gas pressure is much less than the atmospheric pressure.
*This could be managed by keeping the gas in a cylinder with a piston on the bottom supporting a constant load that hangs from the piston.
*Now, the cylinder is put into the warm environment to expend the gas.
*Using the ideal gas equation, we can find the temperature.
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