Two cylinders \(\mathrm{A}\) and \(\mathrm{B}\) fitted with piston contain equal amounts of an ideal diatomic gas at \(300 \mathrm{k}\). The piston of \(\mathrm{A}\) is free to move, While that of \(B\) is held fixed. The same amount of heat is given to the gas in each cylinders. If the rise in temperature of the gas in \(\mathrm{A}\) is \(30 \mathrm{~K}\), then the rise in temperature of the gas in \(\mathrm{B}\) is. (A) \(30 \mathrm{~K}\) (B) \(42 \mathrm{~K}\) (C) \(18 \mathrm{~K}\) (D) \(50 \mathrm{~K}\)

Short Answer

Expert verified
The rise in temperature of the gas in cylinder B is the same as in cylinder A, which is \(30 K\). (Option A)

Step by step solution

01

Calculate the change in internal energy of gas in cylinder A

To find the change in internal energy, we use the formula: \[ \Delta U = n C_v \Delta T \] where \(\Delta U\) is the change in internal energy, \(n\) is the number of moles, \(C_v\) is the molar specific heat at constant volume, and \(\Delta T\) is the change in temperature. For an ideal diatomic gas, we have the following values: - \(C_v = \frac{5}{2} R\), where \(R\) is the universal gas constant Now, the change in temperature for cylinder A is given as \(\Delta T_A = 30\) K. The change in internal energy or \(\Delta U_A\) will be: \[ \Delta U_A = n C_v \Delta T_A = n\left(\frac{5}{2} R\right)(30 K) \]
02

Calculate the work done by gas in cylinder A

Using the first law of thermodynamics, we have: \[ q_A = \Delta U_A + W_A \] where \(q_A\) is the heat given to cylinder A, \(\Delta U_A\) is the change in internal energy, and \(W_A\) is the work done by the gas in cylinder A. We know the heat given to both cylinders is equal. So, we have: \[ W_A = q_A - \Delta U_A \]
03

Calculate the change in internal energy of gas in cylinder B

Since the piston in cylinder B is held fixed, no work is done, and we have: \[ \Delta U_B = q_B \] As the heat given to both cylinders is equal, we get: \[ \Delta U_B = q_B = q_A \]
04

Calculate the rise in temperature of gas in cylinder B

We know that \(\Delta U_B = n C_v \Delta T_B\). We can use this to find the change in temperature for cylinder B, denoted as \(\Delta T_B\): \[ \Delta T_B = \frac{\Delta U_B}{nC_v} = \frac{q_A}{n\left(\frac{5}{2}R\right)} \] Since \(q_A = \Delta U_A + W_A\), we can substitute this in the formula for \(\Delta T_B\): \[ \Delta T_B = \frac{\Delta U_A + W_A}{n\left(\frac{5}{2}R\right)} = \frac{n\left(\frac{5}{2}R\right)(30 K) + W_A}{n\left(\frac{5}{2} R\right)} \] Simplifying, we get: \[ \Delta T_B = 30 K + \frac{W_A}{n\left(\frac{5}{2}R\right)} \]
05

Find the value of \(\Delta T_B\)

Now we need to determine the exact value of \(\Delta T_B\). First, we need to find the value of \(W_A\). From step 2, we have: \[ W_A = q_A - \Delta U_A = q_A - n\left(\frac{5}{2} R\right)(30 K) \] Since \(q_A = \Delta U_B\), we have: \[ W_A = \Delta U_B - n\left(\frac{5}{2} R\right)(30 K) = n\left(\frac{5}{2} R\right)(\Delta T_B - 30 K) \] Plugging this into the formula for \(\Delta T_B\): \[ \Delta T_B = 30 K + \frac{n\left(\frac{5}{2} R\right)(\Delta T_B - 30 K)}{n\left(\frac{5}{2} R\right)} \] Which simplifies to: \[ \Delta T_B = 30 K + (\Delta T_B - 30 K) \] This gives us: \[ \Delta T_B = 30 K \] So the rise in temperature of the gas in cylinder B is the same as in cylinder A, which is 30 K (Option A).

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Most popular questions from this chapter

An ideal refrigerator has a freezer at a temperature of \(-13\) C, The coefficient of performance of the engine is 5 . The temperature of the air to which heat is rejected will be. (A) \(325^{\circ} \mathrm{C}\) (B) \(39^{\circ} \mathrm{C}\) (C) \(325 \mathrm{~K}\) (D) \(320^{\circ} \mathrm{C}\)

An insulated contains containing monoatomic gas of moles mass \(\mathrm{M}_{0}\) is moving with a velocity, \(\mathrm{V}\). If the container is suddenly stopped, find the change in temperature. (A) \(\left\\{\left(M_{0} V^{2}\right) /(5 R)\right\\}\) (B) $\left\\{\left(\mathrm{M}_{0} \mathrm{~V}^{2}\right) /(4 \mathrm{R})\right\\}$ (C) $\left\\{\left(\mathrm{M}_{0} \mathrm{~V}^{2}\right) /(3 \mathrm{R})\right\\}$ (D) $\left\\{\left(\mathrm{M}_{0} \mathrm{~V}^{2}\right) /(2 \mathrm{R})\right\\}$

If \(r\) denotes the ratio of adiabatic of two specific heats of a gas. Then what is the ratio of slope of an adiabatic and isothermal $\mathrm{P} \rightarrow \mathrm{V}$ curves at their point of intersection ? (A) \((1 / \gamma)\) (B) \(\gamma-1\) (C) \(\gamma\) (D) \(\gamma+1\)

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Water of volume 2 liter in a container is heated with a coil of $1 \mathrm{kw}\( at \)27^{\circ} \mathrm{C}$. The lid of the container is open and energy dissipates at the rate of \(160(\mathrm{~J} / \mathrm{S}) .\) In how much time temperature will rise from \(27^{\circ} \mathrm{C}\) to $77^{\circ} \mathrm{C}\(. Specific heat of water is \)4.2\\{(\mathrm{KJ}) /(\mathrm{Kg})\\}$ (A) \(7 \mathrm{~min}\) (B) \(6 \min 2 \mathrm{~s}\) (C) \(14 \mathrm{~min}\) (D) \(8 \min 20 \mathrm{~S}\)

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