1. ⇒ (MHT CET 2023 13th May Morning Shift )

$dQ$ is the heat energy supplied to an ideal gas under isochoric conditions. If $dU$ and $dW$ denote the change in internal energy and the work done respectively then

A. $dQ = dW$

B. $dQ > dU$

C. $dQ < dU$

D. $dQ = dU$

Correct Option is (D)

Under isochoric (constant volume) conditions, the heat energy (dQ) supplied to an ideal gas contributes solely to the change in its internal energy (dU).

Mathematically, this can be expressed as dQ = dU + dW

$\Rightarrow d Q = d U$ ( $∵ dW = 0$ )

2. ⇒ (MHT CET 2023 12th May Evening Shift )

Select the WRONG statement from the following. For an isothermal process

A. Energy exchanged is used to do work

B. Perfect thermal equilibrium with environment

C. Equation of state PV is not constant.

D. No change internal energy.

Correct Option is (C)

Currently no explanation available

3. ⇒ (MHT CET 2023 12th May Evening Shift )

A monoatomic ideal gas initially at temperature ' $T_{1}$ ' is enclosed in a cylinder fitted with massless, frictionless piston. By releasing the piston suddenly the gas is allowed to expand to adiabatically to a temperature ' $T_{2}$ '. If ' $L_{1}$ ' and ' $L_{2}$ ' are the lengths of the gas columns before and after expansion respectively, then $\frac{T_{2}}{T_{1}}$ is

A. $\frac{L_{1}}{L_{2}}$

B. $\frac{L_{2}}{L_{1}}$

C. ${(\frac{L_{1}}{L_{2}})}^{2 / 3}$

D. ${(\frac{L_{2}}{L_{1}})}^{2 / 3}$

Correct Option is (C)

$\begin{array}{ll} For an adiabatic process \\ T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1} \\ ∴ & \frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} \end{array}$

For a monoatomic gas, $r = \frac{5}{3}$

$\begin{array}{cl} \Rightarrow γ - 1 = \frac{5}{3} - 1 = \frac{2}{3} \\ V_{1} = A L_{1} and V_{2} = A L_{2} \\ ∴ & \frac{T_{2}}{T_{1}} = {[\frac{A L_{1}}{A L_{2}}]}^{2 / 3} = {[\frac{L_{1}}{L_{2}}]}^{2 / 3} \\ γ - 1 = \frac{2}{3} \end{array}$

For an adiabatic process,

$\begin{aligned} \frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} = {(\frac{V_{1}}{V_{2}})}^{\frac{2}{3}} \\ V \propto L \\ \frac{T_{2}}{T_{1}} = {(\frac{L_{1}}{L_{2}})}^{\frac{2}{3}} \end{aligned}$

4. ⇒ (MHT CET 2023 12th May Morning Shift )

A gas at normal temperature is suddenly compressed to one-fourth of its original volume. If $\frac{C_{p}}{C_{v}} = γ = 1.5$ , then the increase in its temperature is

A. 273 K

B. 373 K

C. 473 K

D. 573 K

Correct Option is (A)

Given that, $V_{2} = \frac{V_{1}}{4}, \frac{C_{p}}{C_{v}} = γ = 1.5$

As the process is sudden, it is an adiabatic expansion,

$\begin{aligned} ∴ T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1} \\ ∴ T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1} \\ = T_{1} (4)^{γ - 1} \\ = T_{1} \times (4)^{0.5} \\ = 2 T_{1} \\ T_{2} - T_{1} = T_{1} \\ T_{2} - T_{1} = 273 K (∵ T_{1} = Normal temperature) \end{aligned}$

5. ⇒ (MHT CET 2023 11th May Evening Shift )

An ideal gas expands adiabatically. $(γ = 1 \cdot 5)$ To reduce the r.m.s. velocity of the molecules 3 times, the gas has to be expanded

A. 81 times

B. 27 times

C. 9 times

D. 3 times

Correct Option is (A)

We know,

$\begin{aligned} V_{rms} = \sqrt{\frac{3 R T}{M_{0}}} \\ \Rightarrow T \propto V^{2} \\ \frac{T_{2}}{T_{1}} = \frac{V_{2}^{2}}{V_{1}^{2}} = \frac{{(\frac{V_{1}}{3})}^{2}}{V_{1}^{2}} = \frac{1}{9} .... (i) \\ Also, T V^{γ - 1} = Constant \\ \Rightarrow \frac{V_{2}}{V_{1}} = {(\frac{T_{1}}{T_{2}})}^{\frac{1}{γ - 1}} \\ \frac{V_{2}}{V_{1}} = (9)^{2} = 81 \end{aligned}$

⇒Thermodynamics