Answer
Verified
385.8k+ views
Hint To solve this question, we need to use the first law of thermodynamics. For that, we have to determine the work done by using its expression for the isobaric process. Then substituting the value of the work done, and the given value of the heat, we will get the value of the change in the internal energy.
Formula used: The formulae used to solve this question are
${W_p} = P\Delta V$, here ${W_p}$ is the work done in an isobaric process by a pressure of $P$ which changes the volume by $\Delta V$.
$Q = \Delta U + W$, here $Q$ is the heat, $\Delta U$ is the change in the internal energy, and $W$ is the work done.
Complete step-by-step solution:
Since the pressure acting on the gas is constant, so the gas undergoes an isobaric process in which the work done is given by
${W_p} = P\Delta V$
$ \Rightarrow {W_p} = P\left( {{V_2} - {V_1}} \right)$
According to the question, $P = 4.5 \times {10^5}Pa$, ${V_1} = 0.5{m^3}$ and ${V_2} = 2.0{m^3}$. Substituting these in above, we get
\[{W_p} = 4.5 \times {10^5}\left( {2.0 - 0.5} \right)\]
$ \Rightarrow {W_p} = 4.5 \times {10^5} \times 1.5$
On solving we get
${W_p} = 6.75 \times {10^5}J$
The heat given to the gas is equal to $800KJ$. So we have $Q = 800KJ$.
We know that $1KJ = {10^3}J$. So we get
$Q = 800 \times {10^3}J$
$ \Rightarrow Q = 8 \times {10^5}J$
From the first law of thermodynamics, we have
$Q = \Delta U + W$
Substituting $Q = 8 \times {10^5}J$ and $W = {W_p} = 6.75 \times {10^5}J$ in the above equation, we get
$8 \times {10^5} = \Delta U + 6.75 \times {10^5}$
$ \Rightarrow \Delta U = 8 \times {10^5} - 6.75 \times {10^5}$
On solving we finally get
$\Delta U = 1.25 \times {10^5}J$
Thus, the change in internal energy of the gas is equal to $1.25 \times {10^5}J$.
Hence, the correct answer is option D.
Note: The sign convention for the heat must be carefully followed. Since the system is receiving the heat, it is taken as positive. Also, do not forget to write the value of the heat in Joules, which is given in kiloJoules.
Formula used: The formulae used to solve this question are
${W_p} = P\Delta V$, here ${W_p}$ is the work done in an isobaric process by a pressure of $P$ which changes the volume by $\Delta V$.
$Q = \Delta U + W$, here $Q$ is the heat, $\Delta U$ is the change in the internal energy, and $W$ is the work done.
Complete step-by-step solution:
Since the pressure acting on the gas is constant, so the gas undergoes an isobaric process in which the work done is given by
${W_p} = P\Delta V$
$ \Rightarrow {W_p} = P\left( {{V_2} - {V_1}} \right)$
According to the question, $P = 4.5 \times {10^5}Pa$, ${V_1} = 0.5{m^3}$ and ${V_2} = 2.0{m^3}$. Substituting these in above, we get
\[{W_p} = 4.5 \times {10^5}\left( {2.0 - 0.5} \right)\]
$ \Rightarrow {W_p} = 4.5 \times {10^5} \times 1.5$
On solving we get
${W_p} = 6.75 \times {10^5}J$
The heat given to the gas is equal to $800KJ$. So we have $Q = 800KJ$.
We know that $1KJ = {10^3}J$. So we get
$Q = 800 \times {10^3}J$
$ \Rightarrow Q = 8 \times {10^5}J$
From the first law of thermodynamics, we have
$Q = \Delta U + W$
Substituting $Q = 8 \times {10^5}J$ and $W = {W_p} = 6.75 \times {10^5}J$ in the above equation, we get
$8 \times {10^5} = \Delta U + 6.75 \times {10^5}$
$ \Rightarrow \Delta U = 8 \times {10^5} - 6.75 \times {10^5}$
On solving we finally get
$\Delta U = 1.25 \times {10^5}J$
Thus, the change in internal energy of the gas is equal to $1.25 \times {10^5}J$.
Hence, the correct answer is option D.
Note: The sign convention for the heat must be carefully followed. Since the system is receiving the heat, it is taken as positive. Also, do not forget to write the value of the heat in Joules, which is given in kiloJoules.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Why Are Noble Gases NonReactive class 11 chemistry CBSE
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE