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Question: A gas under constant pressure of \(4.5 \times {10^5}Pa\) when subjected to \(800KJ\) of heat, change...

A gas under constant pressure of 4.5×105Pa4.5 \times {10^5}Pa when subjected to 800KJ800KJ of heat, changes the volume from 0.5m30.5{m^3} to 2.0m32.0{m^3}. The change in the internal energy of the gas:
(A) 6.75×105J6.75 \times {10^5}J
(B) 5.25×105J5.25 \times {10^5}J
(C) 3.25×105J3.25 \times {10^5}J
(D) 1.25×105J1.25 \times {10^5}J

Explanation

Solution

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
Wp=PΔV{W_p} = P\Delta V, here Wp{W_p} is the work done in an isobaric process by a pressure of PP which changes the volume by ΔV\Delta V.
Q=ΔU+WQ = \Delta U + W, here QQ is the heat, ΔU\Delta U is the change in the internal energy, and WW 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
Wp=PΔV{W_p} = P\Delta V
Wp=P(V2V1)\Rightarrow {W_p} = P\left( {{V_2} - {V_1}} \right)
According to the question, P=4.5×105PaP = 4.5 \times {10^5}Pa, V1=0.5m3{V_1} = 0.5{m^3} and V2=2.0m3{V_2} = 2.0{m^3}. Substituting these in above, we get
Wp=4.5×105(2.00.5){W_p} = 4.5 \times {10^5}\left( {2.0 - 0.5} \right)
Wp=4.5×105×1.5\Rightarrow {W_p} = 4.5 \times {10^5} \times 1.5
On solving we get
Wp=6.75×105J{W_p} = 6.75 \times {10^5}J
The heat given to the gas is equal to 800KJ800KJ. So we have Q=800KJQ = 800KJ.
We know that 1KJ=103J1KJ = {10^3}J. So we get
Q=800×103JQ = 800 \times {10^3}J
Q=8×105J\Rightarrow Q = 8 \times {10^5}J
From the first law of thermodynamics, we have
Q=ΔU+WQ = \Delta U + W
Substituting Q=8×105JQ = 8 \times {10^5}J and W=Wp=6.75×105JW = {W_p} = 6.75 \times {10^5}J in the above equation, we get
8×105=ΔU+6.75×1058 \times {10^5} = \Delta U + 6.75 \times {10^5}
ΔU=8×1056.75×105\Rightarrow \Delta U = 8 \times {10^5} - 6.75 \times {10^5}
On solving we finally get
ΔU=1.25×105J\Delta U = 1.25 \times {10^5}J
Thus, the change in internal energy of the gas is equal to 1.25×105J1.25 \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.