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Chapter 8: Problem 1099

In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas release \(20 \mathrm{~J}\) of heat and $8 \mathrm{~J}$ of work has done on the gas-If the initial internal energy of the gas was \(30 \mathrm{j}\), then the final internal energy will be (A) \(58 \mathrm{~J}\) (B) \(2 \mathrm{~J}\) (C) \(42 \mathrm{~J}\) (D) \(18 \mathrm{~J}\)

Short Answer

Expert verified
The final internal energy of the gas is \(18 \mathrm{~J}\), which is option (D).

Step by step solution

01

Write down the given values

We are given the following values: Initial internal energy (U₁) = 30 J Heat released (Q) = -20 J (Since heat is released, it is negative) Work done on the gas (W) = 8 J (Since work is done on the gas, it is positive)
02

Apply the first law of thermodynamics

Using the first law of thermodynamics, we can find the change in internal energy (ΔU): ΔU = Q + W
03

Substitute the given values

We can now substitute the given values into the equation: ΔU = (-20 J) + (8 J)
04

Calculate the change in internal energy

Now, we can calculate the change in internal energy: ΔU = -12 J
05

Find the final internal energy

To find the final internal energy (U₂), we will use the equation: U₂ = U₁ + ΔU Substitute the given values: U₂ = (30 J) + (-12 J)
06

Calculate the final internal energy

Finally, calculate the final internal energy: U₂ = 18 J The final internal energy (U₂) is 18 J, which corresponds to option (D).

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Most popular questions from this chapter

In Column I different Process is given match corresponding option of column \(\mathrm{I}_{1}\) Column - I Columm - II (A) adiabatic process (p) \(\Delta \mathrm{p}=0\) (B) Isobaric process (q) \(\Delta \mathrm{u}=0\) (C) Isochroic process (r) \(\Delta Q=0\) (D) Isothermal process (s) \(\Delta \mathrm{W}=0\) (A) A-p, B-s, C-r, D-q (B) (B) A-s, B-q, C-p, D-r (C) $\mathrm{A}-\mathrm{r}, \mathrm{B}-\mathrm{p}, \mathrm{C}-\mathrm{s}, \mathrm{D}-\mathrm{q}$ (D) $\mathrm{A}-\mathrm{q}, \mathrm{B}-\mathrm{r}, \mathrm{C}-\mathrm{q}, \mathrm{D}-\mathrm{p}$

A diatomic gas initially at \(18^{\circ} \mathrm{C}\) is Compressed adiabatically to one eighth of its original volume. The temperature after Compression will be (A) \(10^{\circ} \mathrm{C}\) (B) \(668 \mathrm{~K}\) (C) \(887^{\circ} \mathrm{C}\) (D) \(144^{\circ} \mathrm{C}\)

One \(\mathrm{kg}\) of adiatomic gas is at a pressure of $5 \times 10^{5}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$ The density of the gas is \(\left\\{(5 \mathrm{~kg}) / \mathrm{m}^{3}\right\\}\) what is the energy of the gas due to its thermal motion ? (A) \(2.5 \times 10^{5} \mathrm{~J}\) (B) \(3.5 \times 10^{5} \mathrm{~J}\) (C) \(4.5 \times 10^{5} \mathrm{~J}\) (D) \(1.5 \times 10^{5} \mathrm{~J}\)

A Carnot engine operating between temperature \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) has efficiency \(0.4\), when \(\mathrm{T}_{2}\) lowered by $50 \mathrm{~K}\(, its efficiency increases to \)0.5\(. Then \)\mathrm{T}_{1}$ and \(\mathrm{T}_{2}\) are respectively. (A) \(300 \mathrm{~K}\) and \(100 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) and \(200 \mathrm{~K}\) (C) \(600 \mathrm{~K}\) and \(400 \mathrm{~K}\) (D) \(500 \mathrm{~K}\) and \(300 \mathrm{~K}\)

The temperature of a substance increases by \(27^{\circ} \mathrm{C}\) What is the value of this increase of Kelvin scale? (A) \(300 \mathrm{~K}\) (B) \(2-46 \mathrm{~K}\) (C) \(7 \mathrm{~K}\) (D) \(27 \mathrm{~K}\)
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